An Introduction to pH and Buffer Solutions Calculations [work in progress] Code (Text): This is a work in progress 1. pH and Strong Acids pH is defined as the negative logarithm of the hydrogen ion concentration in a solution; pH = - log_{10}[H^{+}(aq)]. (From this point forward 'log' will represent log to the base ten). The square brackets above indicate that it is the concentration we are concerned with. Now, since a strong acid effectively dissociates completely into its respective ions when reacted with water; the amount of moles of protons in solution is equal to the number of moles of H^{+} in the acid. This is not actually the case since some water auto-ionises; however, this effect is insignificant when dealing with acids with a concentration greater than around 5x10^{-7} mol.dm^{-3}. The auto-ionisation of water will be discussed later. One must be careful here to note whether a monoprotic or diprotic acid is used. A monoprotic acid has a single ionizable hydrogen atom per acid molecule; a diprotic has two, polyprotic many. Worked Examples Calculate the pH of a solution of Hydrochloric acid with a concentration of 0.01 mol.dm^{-3} First we must write the equation for the dissociation of the acid in water; [tex]HCl(aq) + H_{2}O(l) \rightarrow H_{3}O^{+}(aq) + Cl^{-}(aq)[/tex] We can further simplify this to [tex]HCl(aq) \rightarrow H^{+}(aq) + Cl^{-}(aq)[/tex] The water and hence the oxonium ion can be ignored since the water is present in excess. From the latter equation we can see that Hydrochloric acid is a monoprotic strong acid. Therefore, the concentration of protons^{*} in solution is equal to the concentration of the acid, hence the pH is; [tex]pH = - \log\left[ H^{+}(aq)\right] = - \log\left[ 0.01\right] = 2[/tex] Calculate the pH of a solution of Sulphuric Acid with a concentration of 0.01 mol.dm^{-3} Again we should start by writing an equation for the dissociation; [tex]H_{2}SO_{4}(aq) \rightarrow 2H^{+}(aq) + SO_{4}^{2-}(aq)[/tex] Note that sulphuric acid is a diprotic acid, therefore to obtain the concentration of protons on solution we must double the concentration of the acid; [tex]pH = - \log\left[ H^{+}(aq)\right] = -\log\left[ 2\times 0.01\right] \approx 1.70[/tex] So although the sulphuric acid and the hydrochloric acid are the same concentration, sulphuric acid has a lower pH since it is diprotic. We can also use the pH to calculate the concentration of a particular acid. A solution of Nitric Acid has a pH of 1.2, determine the concentration of the acid used The first step as with all these calculations is to write the dissociation equation for the acid; [tex]HNO_{3}(aq) \rightarrow H^{+}(aq) + NO_{3}^{-}[/tex] Rearranging the standard equation for pH we obtain; [tex]pH = -\log\left[ H^{+}(aq)\right] \Leftrightarrow \left[ H^{+}(aq)\right] = 10^{-pH}[/tex] [tex]\left[ H^{+}(aq)\right] = 10^{-1.2} \approx 0.0631 mol.dm^{-3}[/tex] Since nitric acid is monoprotic the concentration of hydrogen ions in solution is equal to the concentration of the acid; therfore, the nitric acid has a concentration of 0.0631 mol.dm^{-3} (3sf) _________________________________________ 2.Weak Acids Weak acids do not completely dissociate when in solution, instead an equilibrium is set up; [tex]HA(aq)\rightleftharpoons H^{+}(aq) + A^{-}(aq)[/tex] We can write an equilibrium expression for this reaction. In this case the constant is known as the acidity constant or acid dissociation constant and is given the symbol K_{a}; [tex]K_{a} = \frac{\left[ H^{+}(aq)\right] \left[ A^{-}(aq) \right]}{\left[ HA(aq) \right]}[/tex] Here, we can now make two assumptions which will grately simplify the calculations; Assumption One: [H^{+}(aq)] = [A^{-}(aq)] This means that we are ignoring the effect of the ionisation of water, this is a valid assumption for most 'normal' concentrations. Assumption Two: As weak acids dissociate very little, we can assume that the concentration of the acid at equilibrium is the same as if the acid did not dissociate With these assumptions we can rewrite the equilibrium constant as follows; [tex]K_{a} = \frac{\left[ H^{+}(aq)\right]^2}{\left[ HA(aq) \right]}[/tex] The K_{a} values are spread of a large range, therefore it is often more convenient to define the term pK_{a} = - log_{10} K_{a} Worked Example Knowing that K_{a} for ethanoic acid is 1.7x10^{-5} mol.dm^{-3} at 298K, calculate the pH of a solution of enthanoic acid with a concentration of 1 mol.dm^{-3} at 298K. The temperature is important here as it effects the position of the equilibrium by Le chatelier's principle, therefore K_{a} will be affected. Rearranging our equation we have: [tex]K_{a} = \frac{\left[ H^{+}(aq)\right]^2}{\left[ HA(aq) \right]} \Leftrightarrow \left[ H^{+}(aq)\right] = \sqrt{K_{a} \cdot \left[ HA(aq) \right]}[/tex] Thus: [tex]pH = -\log \left( \sqrt{K_{a} \cdot \left[ HA(aq) \right]} \right) = -\log \left( \sqrt{(1.7\times10^{-5}) \cdot [ 1 ]} \right)[/tex] [tex]pH \approx 2.38 (\text{at 289K})[/tex] _________________________________________ 3. Auto-Ionisation of Water Although water is not considered ionic, it does dissociate to some extent. Water is amphoretic (able to act both as a base and an acid) and will auto-dissociate into the oxonium ion H_{3}O^{+} and the hydroxide ion; [tex]2H_{2}O(l) \rightleftharpoons H_{3}O^{+}(aq) + OH^{-}(aq)[/tex] We can write now write an expression for the equilibrium constant, which in this case is known as the ionic product of water (K_{w}); [tex]K_{w} = \frac{\left[ H_{3}O^{-}(aq) \right] \left[OH^{-}(aq) \right]}{[H_{2}O(l)]}[/tex] In pH calculations we can treat the oxonium ion as a proton, and since the water is present in excess the [H_{2}O_{(l)}] term is effectively constant and therefore can be omitted; thus the K_{w} expession becomes; [tex]K_{w} = \left[ H^{+}(aq) \right] \left[OH^{-}(aq) \right][/tex] This expression will be important when dealing with bases Water has a K_{w} value of 1 x 10^{-14} mol^{2}.dm^{-6 at 298K (note the units here, since we are multiplying two concentrations together we must square the units), hence the pH of water at 298K is; [tex]K_{w} = \left[ H^{+}(aq) \right] \left[OH^{-}(aq) \right][/tex] In pure water [H+(aq)] = [OH-(aq)], therefore; [tex] \left[ H^{+}(aq) \right] = \sqrt{K_{w}}[/tex] [tex]pH = - \log \left| \sqrt{k_{w}} \right| = -\log \left| \sqrt{1\times 10^{-14}} \right|[/tex] [tex]pH = 7[/tex] An interesting point to note, is that the auto-dissociation of water is an endothermic process, therefore, by Le Chatelier's principle if we increase the temperature the position of the equilibrium will shift to oppose that change, the value of Kw will decrease and therefore the pH of water falls as we increase the temperature. _________________________________________* Note that a H+ is a proton. Both terms will be used interchangeably here Code (Text): Sections to Follow; [LIST] [*]pH of Strong Bases [*]Weak Bases [*]Buffer Solutions, applications, mechanism and Calculations [*]Advance treatments taking into account the autoionisation of water [/LIST] Please feel free to PM with any suggestions, comments or corrections. }
4. pH of Strong Bases The expression for the ionic product of water (K_{w}) above can be used to determine the pH (or molarity) of a basic solution. Again, the calculations are greatly simplified if we assume that the base completely dissociates when placed in solution; [tex]BOH(aq) \rightarrow B^{+}(aq) + OH^{-}(aq)[/tex] In this case; [tex]\left[ OH^{-}(aq) \right] = \left[ BOH(aq) \right][/tex] Using the expression of the ionic product of water; [tex]K_{w} = \left[ H^{+}(aq) \right]\left[ OH^{-}(aq) \right] \Leftrightarrow \left[ H^{+}(aq) \right] = \frac{K_{w}}{\left[ OH^{-}(aq) \right] }[/tex] Thus the pH is given by; [tex]pH = -\log(\frac{K_{w}}{\left[ OH^{-}(aq) \right] })[/tex] However, even the strongest bases are weak when compared to strong acids (they do not have a strong tendency to dissociate). For example, the pK_{b} for Potassium Hydroxide is only around 0.5. This leads to calculative errors with concentrated basic solutions. _________________________________________
Hey, this is a great thread! I always wanted to do something like this, cause there are many such questions in the homework forums. Do you mind if I contribute? I can help in mixtures such as SA-WB, WA-SB, WA-WB, buffers, etc if you want.
Thank you for your compliment Feel free to add whatever you wish, I'm going to be pretty busy for the next week or so, so I won't have much time to contribute as much as I would wish. I'm probably going add a few worked examples to my post on strong bases tomorrow, but aside from that feel free to contribute (and point out any corrections) anything.
5. pH of Buffer Solutions What are Buffer solutions? A Buffer solution resists changes in pH on dilution or when small quantities of an acid or a base is added to it. Buffer solutions are generally prepared by mixing Weak acid and its salt (Ex:CH3COOH and CH3COONa) Weak base and its salt (Ex:NH4OH and NH4Cl) So, buffers contain either a weak acid and its conjugate base, or a weak base and its conjugate acid. Henderson-Hasselbalch equations Now, an important question is, how do we find the pH of a buffer solution? Acid Buffer solutions First, let's assume that the buffer solution is made of a weak acid HA (such as, CH3COOH) and its conjugate base A^{-} (such as, CH3COO^{-}). Now, look at the ionisation of HA in the buffer solution. As already discussed in the first post, weak acids do not completely dissociate, and instead an equilibrium is set up. [tex]HA(aq)\rightleftharpoons H^{+}(aq) + A^{-}(aq)[/tex] By the Le Chatelier's principle, if we have lots of the conjugate base A^{-}, the position of the equilibrium will be to the left. In this particular case, there are some assumptions we make. Assumption 1: All the salt AB present, dissociates almost completely into its ions A^{-} and B^{+}. So, there's a lot of A^{-} from the salt, and as a consequence, very little of the weak acid HA dissociates. Now consider the equilibrium expression for this reaction. [tex]K_{a} = \frac{\left[ H^{+}(aq)\right] \left[ A^{-}(aq) \right]}{\left[ HA(aq) \right]}[/tex] which is rearranged to give [tex]H^{+}(aq)= \frac{K_{a} \left[ HA(aq) \right]}{ \left[ A^{-}(aq) \right]}[/tex] and taking the negative logarithm on both sides, [tex]pH = pK_{a} + \log(\frac{\left[A^{-}(aq)\right]}{\left[HA(aq) \right]})[/tex] The above equation is known as the Henderson-Hasselbalch equation for acid buffer solutions. In the above equation, As per our assumption, the amount of A^{-} present is mainly due to the salt AB. So, [A^{-}(aq)] = [AB(aq)]. This means that, the amount of A^{-} present at equilibrium is approximately the same as the amount of salt AB present initially. Similarly, very little of the acid undergoes dissociation. Therefore, at equilibrium, the amount of undissociated acid is approximately the same as the amount of acid which was present initially. As a result, the Henderson-Hasselbalch equation is also written as [tex] pH = pK_{a} +\log(\frac{[salt]}{[acid]}) [/tex] where [itex][salt][/itex] and [itex][acid][/itex] refer to what was present initially.
6. pH of Weak Bases Weak bases do not fully dissociate when placed in solution, an equilibrium is obtained similar to weak acids; [tex]BOH(aq) \rightleftharpoons B^{+}(aq) + OH^{-}(aq)[/tex] As with the acidic equilibrium, we can defined an equilibrium constant, which in this case is known as the basicity constant (K_{b}); [tex]K_{b} = \frac{\left[ B^{+}(aq)\right] \left[ OH^{-}(aq) \right]}{\left[ BOH(aq) \right]}[/tex] Again, here we can make a number of assumptions which greatly simplify the calculations. These assumptions hold when the base is not extremely weak or very dilute (i.e. we can ignore the ionic product of water). These assumptions are as follows and are similar to the assumptions made when calculating the pH of weak acids; Assumption One: [B^{+}(aq)] = OH^{-}(aq)] This means that we are ignoring the effect of the ionisation of water, this is a valid assumption for most 'normal' concentrations. Assumption Two: As weak bases dissociate very little, we can assume that the concentration of the base at equilibrium is the same as if the base did not dissociate at all. These assumptions allows us to rewrite the basicity constant as follows; [tex]K_{b} = \frac{\left[ OH^{-}(aq) \right]^{2}}{\left[ BOH(aq) \right]}[/tex] An expression for pH can then be found thus; [tex]K_{b} = \frac{\left[ OH^{-}(aq) \right]^{2}}{\left[ BOH(aq) \right]} \Leftrightarrow \left[ OH^{-}(aq) \right] = \sqrt{K_{b} \cdot \left[ BOH(aq) \right]}[/tex] Using the expression for the ionic product of water; [tex]K_{w} = \left[ H^{+}(aq) \right] \left[OH^{-}(aq) \right] \Leftrightarrow \left[OH^{-}(aq) \right] = \frac{K_{w}}{\left[ H^{+}(aq) \right]}[/tex] [tex]\frac{K_{w}}{\left[ H^{+}(aq) \right]} = \sqrt{K_{b} \cdot \left[ BOH(aq) \right]}[/tex] [tex]\left[ H^{+}(aq) \right] = \frac{K_{w}}{\sqrt{K_{b} \cdot \left[ BOH(aq) \right]}}[/tex] [tex]pH = -\log\left( \frac{K_{w}}{\sqrt{K_{b} \cdot \left[ BOH(aq) \right]}} \right)[/tex] Using the laws of logarithms and knowing that pK_{w} = 14 we can show that; [tex]pH = 14 + \log\left( \sqrt{K_{b} \cdot \left[ BOH(aq) \right]}\right)[/tex] Alternatively, one could take the equation for the ionisation product of water and find in the pH from the pOH of the basic solution; [tex]\left[ OH^{-}(aq) \right] = \sqrt{K_{b} \cdot \left[ BOH(aq) \right]} \Leftrightarrow pOH = -\log\sqrt{K_{b} \cdot \left[ BOH(aq) \right]}[/tex] [tex]K_{w} = \left[ H^{+}(aq) \right] \left[OH^{-}(aq) \right] \Leftrightarrow -\log K_{w} = pH \times pOH[/tex] Using the laws of logarithms; [tex]pH = -\log (1\times10^{-14}) - pOH[/tex] [tex]pH = 14 + \log\left( \sqrt{K_{b} \cdot \left[ BOH(aq) \right]}\right)[/tex]
[tex]K_{w} = \left[ H^{+}_{(aq)} \right] \left[OH^{-}_{(aq)} \right][/tex] It follows automatically that, when [tex]K_{W} = 1.0 \cdot 10^{-14}[/tex], [tex]pH + pOH = 14[/tex], and more general: [tex] pH + pOH = pK_{W}[/tex]
Thanks Many thanks for this great tutorial. I would also suggest that you discuss some tricky questions on acids/bases and pH like: 1.How a solution of Ammonium Acetate can act as a buffer 2.What is the pH of 10^-8 m HCL solotion? 3. Can pH be lesser than zero or greater than 14? Thanks again D_R
7. General Case The above described methods for finding the pH of acids, bases and buffers all make some assumptions. Usually, when dealing with solutions that are within the 'normal range' of pH these assumptions are not significant. However, when we reach some extremes, these assumptions indeed become significant. When this is the case we must disregard all these assumptions and take a general approach. Once such approach is outlined here. Note, that this approach (for the sake of lucidity) assumes that the acid is monoprotic. It was previously said in this thread that a strong acid in solution is described as fully dissociated, however, even in the case of strong acids / bases this is not the case. Therefore, a solution of a strong acid (HA) exists at equilibrium and can be represented by two equilibrium equations; This represents the acidity constant (K_{a}) of the acid (Eq.1); [tex]K_{a} = \frac{\left[ H^{+}(aq)\right] \left[ A^{-}(aq) \right]}{\left[ HA(aq) \right]}[/tex] This equation represents the auto-ionisation of water as described in section three above (Eq.2); [tex]K_{w} = \left[ H^{+}(aq) \right] \left[OH^{-}(aq) \right][/tex] The following to expressions describe the conservation of mass and charge The sum of the negative charges must be equal to the hydrogen ion concentration (Eq.3); [tex]\left[ H^{+}(aq) \right] = \left[OH^{-}(aq) \right] + \left[A^{-}(aq) \right][/tex] Conservation of mass implies that the sum of the concentration of all forms of the acid must be equal to the concentration of the acid added (C_{a}) (Eq.4); [tex]\left[ A^{-}(aq) \right] + \left[HA(aq) \right] = C_{a}[/tex] Now, using the above formulae we can generate an equation which can be used to calculate the hydrogen ion concentration of a solution of any acid; strong, weak, dilute, concentrated and any in between. First taking Eq.2 and Eq.3 and combing them to eliminate the [OH^{-}] term, we obtain; [tex]\left[ H^{+}(aq) \right] - \left[A^{-}(aq) \right] = \frac{K_{w}}{\left[ H^{+}(aq) \right]}[/tex] Next, if we combine Eq.1 and Eq.4 to eliminate we achieve; [tex]K_{a} = \frac{\left[ H^{+}(aq)\right] \left[ A^{-}(aq) \right]}{C_{a} - \left[ A^{-}(aq) \right]}[/tex] Taking the above equation and making [A^{-}] the subject gives; [tex]\left[ A^{-}(aq) \right] = \frac{C_{a}K_{a}}{\left[ H^{+}(aq)\right] + K_{a}}[/tex] Taking this equation and substituting it into the equation we obtained from combining Eq.2 and Eq.3 and re-arranging the equation to be equal to zero we obtain; [tex]\left[ H^{+}(aq)\right] - \frac{K_{w}}{\left[ H^{+}(aq)\right]} - \frac{C_{a}K_{a}}{K + \left[ H^{+}(aq)\right]} = 0[/tex] The above result can be expanded, thus obtaining a cubic equation for the hydrogen ion concentration; [tex]\left[ H^{+}(aq)\right]^{3} + K_{a}\left[ H^{+}(aq)\right]^{2} - (C_{a}K_{a} + K_{w})\left[ H^{+}(aq)\right] - K_{a}K_{w} = 0[/tex] Therefore, if we know the original concentration of the (monoprotic) acid and its dissociation constant we can accurately calculate the pH of a solution of the acid. However, as can be seen from the above equation finding the hydrogen ion concentration exactly can be complicated and very tedious, this is why some assumptions are made. One can also use the above process to find an expression for the concentration of hydroxide groups of a basic solution. However, at the first stage when combining Eq.2 and Eq.3, one would eliminate [H^{+}], rather than [OH^{-}] _________________________________________Once again, I welcome your comments, corrections and suggestions; either via PM or here.
I have a weak acid question Hi, i have a question about weak acids that deal with diprotic and tripotic acids. How would you go about determining the pH of those?? Also, say, for sulphuric acid, the dissociation is as follows: H2SO4 --> H^+ + HSO4^- --> 2H^+ + SO4^2- so, for the second dissociation is the formula as follows? Ka = ([H+]^2 * [SO4^2-])/[H2SO4] do you have to square the concentration of the hydrogen ion because in order to balance the equation, 2 H+'s are required. Or, if you're doing a step by step dissociation, do you not square it because you're taking into account the H+ concentration from the previous dissociationg when calculating the amount of H+?? Thanks!! Oh yeah, and one more thing: when you're testing the 5% rule, do you check it with just the [H+] concentration that you found or if there are 2H+'s in the dissociation, do you have to multiply the [H+] by 2 in order to check the 5% rule thing?
I have been teaching pH in H.S. Chemistry for years and years. I like this thread....it's useful. But, does anyone know the meaning of "p" , or what the "p" in pH means or stands for? Several college profs didn't know.
It is pouvoir hydrogene, I believe. Something French along those lines. No, that's not right. This is correct. You will have to do two step by step calculations with two different Ka values. You don't need to multiply the H+ by 2 because the final concentration of H+ is full dissociation at equilibrium.
Meaning of "p" in pH. The Danish Scientist SPL Sorensen first coined the term "pH" in 1909. pH comes from the latin, "pondus hydrogenii", pondus meaning "weight, burden, mass, balance etc". This term was used to describe the negative base 10 logarithm of the concentration of protons in solution, although was quickly changed as the most chemical and biological systems are governed by "hydrogen ion activity", rather than concentration.
I wasn't aware of this thread up to today, but in 2005 I have prepared my own pH calculation lectures. A little bit more comprehensive. Feel free to browse and comment. One important remark on the pH of weak acids/bases presented here: assumption that acid/base is dissociatied only slightly (and HA or BOH equals analytical concetration) works as long, as the dissociation fraction is below 5%. This should be explicitly checked when doing calculations. Now and again I see studens using [tex]\sqrt{c K_a}[/tex] formula without even knowing that it can be wrong.
Accidentally landed here again. Sulfuric acid - although strong - is not strong enough for the assumption that it is completely dissociated. First proton - OK, but not the second one. In this case second one is dissociated only in about 50% and in effect the real pH is closer to 1.85 than to 1.70.