A practice question reads

What is the $\ce{pH}$ of a solution of a weak acid and its conjugate base when the ratio of the concentrations of the conjugate base/weak acid is 1/3? (the weak acid $K_\mathrm{a} = 6.7\times 10^{-5}$)

a. 5.67

b. 4.65

c. 4.17

d. 3.69

e. 0.481

The answer given is d, I would appreciate it if someone would be wiling to walk me through the steps to solve this problem as my exam is tomorrow morning. Thank you in advance.

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    $\begingroup$ So what is the mathematical formula for the Ka and the various species? $\endgroup$ – MaxW Apr 3 '19 at 18:23
  • $\begingroup$ Sorry, I figured it out. I used $pH = pKa + log([A-]/[HA])$ to get $pH = -log(Ka) + log(1/3)$. $\endgroup$ – Codeblockz Apr 3 '19 at 19:54
  • 2
    $\begingroup$ No reason to be sorry. I was trying to get you to figure it out. That way you learned something. $\endgroup$ – MaxW Apr 3 '19 at 20:10
  • 1
    $\begingroup$ Your comment about the formula for Ka is what reminded me to use the Henderson Hasselbalch equation so thank you. The given ratio was also confusing me even though you just simply plug it into the equation. $\endgroup$ – Codeblockz Apr 3 '19 at 20:16

To solve this I used the Henderson-Hasselbalch equation:

$$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{[\ce{A^-}]}{[\ce{HA}]}\right)$$

Because $\ce{pH} = -\log(\mathrm{p}K_\mathrm{a})$ we end up with:

$\ce{pH} = -\log(\mathrm{p}K_\mathrm{a})+ \log \left(\frac{1}{3}\right)=-\log \left(6.7 \times 10^{-5}\right) + \log \left(\frac{1}{3}\right)= 3.69$.


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