Getting pH of a weak acid and conjugate base with a given ratio [closed]








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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.




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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