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An Introduction to pH and Buffer Solutions Calculations 
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#1
Jun2506, 08:07 AM

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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 autoionises; however, this effect is insignificant when dealing with acids with a concentration greater than around 5x10^{7} mol.dm^{3}. The autoionisation 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 ExamplesWe 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 _________________________________________ 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)]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 _________________________________________ 3. AutoIonisation 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 autodissociate 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[sup]6[sup] 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 autodissociation 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 K_{w} 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



#2
Jul906, 01:24 PM

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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. _________________________________________



#3
Jul906, 03:30 PM

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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 SAWB, WASB, WAWB, buffers, etc if you want.



#4
Jul906, 03:36 PM

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An Introduction to pH and Buffer Solutions Calculations



#5
Jul2306, 02:14 AM

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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
So, buffers contain either a weak acid and its conjugate base, or a weak base and its conjugate acid. HendersonHasselbalch 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 HendersonHasselbalch equation for acid buffer solutions. In the above equation,
As a result, the HendersonHasselbalch 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
Jul2306, 07:13 AM

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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)]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] 


#7
Jul2306, 08:32 AM

P: 92

[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] 


#8
Jul2706, 06:18 AM

P: 2

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 


#9
Aug206, 10:10 AM

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Ok, I'll try adding that and more on buffers over the coming weekend.



#10
Aug706, 01:26 PM

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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;
[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 rearranging 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.



#11
Sep1006, 02:31 PM

P: 1

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? 


#12
Nov2106, 05:24 PM

P: 9

Is this Chemistry 30 from Alberta?



#13
Jan407, 11:45 AM

P: 1

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. 


#14
Jan407, 02:04 PM

P: 262




#15
Apr108, 03:12 AM

P: 1

As far as I know, pH stands for potentia hydrogenii, or power of hydrogen... 


#16
Apr108, 07:25 AM

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#17
Jun1308, 05:03 AM

P: 1




#18
Jun1308, 05:50 AM

Admin
P: 23,724

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. 


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