Calculate the pH of a 0.20 M solution of iodic acid (HIO3).

[JessicaHelena]

Homework Statement

Calculate the pH of a 0.20 M solution of iodic acid $(HIO_3)$. $K_a$ for iodic acid is 0.17

Homework Equations

$K_a = \frac{[H^+][IO_3^-]}{[HIO_3]}$

The Attempt at a Solution

$[HIO_3] = 0.20$ M.

The concentration of $[H^+]$ and $[IO_3^-]$ should be the same because
$HIO_3 + H_2O$ ←→ $H_3O^+ + IO^-$

Therefore, $[H^+][IO_3^-] = 0.20 \times 0.17$, or $[H^+] = \sqrt{0.20 \times 0.17}$. To find the pH, I can do $-\log_{10} [H^+]$, which gives me 0.7343. However, that is apparently wrong.

 

[Charles Link]

You are getting that from this that $[H^+] \approx .185$. You are assuming $[HIO_3]$ in the denominator stays at approximately $[HIO_3] \approx .2$. Clearly this is not the case, and a complete quadratic solution, taking account the drop in $[HIO_3]$ in the denominator is necessary. $\\$ Incidentally, in order to get Latex to work, you need to put # two times on each side of the expression.

 

[JessicaHelena]

I'm unfamiliar with this but this is part of my homework problems... Could you help me understand it better? What quadratic equation should I use and how can I know that [HIO_3] is not 0.2 all the time?

 

[Charles Link]

If $[H^+]=[IO_3^-]=x$ , then $[HIO_3]=.2-x$. $\\$ If the $x$ you had gotten by your method, which ignored the change in the denominator, gave $x << .2$, then $[HIO_3] \approx .2$. But a result like that was found to not be the case. $\\$ For a hint at setting up the quadratic, write the expression for $K_a$ with numerator and denominator both containing the $x$ terms.

 

[epenguin]

You need two of those # where you put one then it comes out better.

Homework Statement

Calculate the pH of a 0.20 M solution of iodic acid $(HIO_3)$. $K_a$ for iodic acid is 0.17

Homework Equations

$K_a = \frac{[H^+][IO_3^-]}{[HIO_3]}$

The Attempt at a Solution

$[HIO_3] = 0.20$ M.

The concentration of $[H^+]$ and $[IO_3^-]$ should be the same because $HIO_3 + H_2O \leftarrow \rightarrow H_3O^+ + IO_3^-$.

Therefore, $[H^+][IO_3^-] = 0.20 \times 0.17$, or $[H^+] = \sqrt{0.20 \times 0.17}$. To find the pH, I can do $-\log_{10} [H^+]$, which gives me 0.7343. However, that is apparently wrong.

 

[epenguin]

How to do the calculation involving the quadratic equation has been explained to you. I hope you can report your calculation and conclusion here.

It would be a bit easier to help you if you had reported slightly more detail of your calculations and reasonings.

You say 'apparently that's wrong' but you do not say why that is apparent. Something you were told?

A couple of tips which I hope will empower you more for calculations in this area, which students generally find it difficult at first to get their minds around as evidenced by the fact that the most frequent single request topic for help in this section of forum is these pH etc. calculations.

Your square root formula for [H⁺] depends on the assumption that the acid is very little dissociated (so [HIO₃] remains near enough to 0.2). If the dissociation were 1% that would be acceptable for most purposes, if it were 10% that would be slightly off but not very bad. But your calculated [H⁺] which equals [IO₃^-] seems to be 0.184 out of molarity 0.2, saying it is more than 90% dissociated! So your result tells you your assumption is self-contradictory - no magic authority about it.

Secondly for this common type of equilibrium when concentrations are around the K_a you get a half-and-half situation. You can see from your equilibrium equation that if [IO₃^-] = [HIO₃] then [H⁺] = K_a (and pH = pK_a). In fact all three species then have equal concentrations here. So if the total concentration of the acid were 0.34 M, [H⁺] would be 0.17. The total acid concentration instead is 0.2 M - in about the same ballpark. (I leave you to think qualitatively do you expect it to be more all less than 50% dissociated?)

My calculation gave [H⁺] = 0.118, pH = 0.93 but I know I I'm quite liable to mistakes. Hope this helps and please conclude when you can (see my sig.)