# Calculate pH of a buffer solution

Calculate the pH in a buffer prepared from $$\pu{50 mL, 0.30 M}$$ formic acid ($$\ce{HCOOH}$$) and $$\pu{30 mL, 0.40 M}$$ sodium formate ($$\ce{HCOONa}$$).

My way of solving:

$$n_{\ce{HCOOH}} = 0.05 \times 0.3 = \pu{0.015 mol}$$

$$n_{\ce{HCOONa}} = 0.03 \times 0.4 = \pu{0.012 mol}$$ (which means $$\ce{NaOH} = \pu{0.012mol}$$)

$$\ce{HCOOH + NaOH <=> HCOONa + H2O}$$

After doing a BCA-table, I got that the remaining moles of HCOOH = $$\pu{0.003 mol}$$

and HCOONa = $$\pu{0.012 mol}$$

Calculating the new concentration:

[HCOOH] = $$\ce{0.003 / 0.08 = 0.0375M}$$

[HCOONa] = $$\ce{0.012 / 0.08 = 0.15M}$$

Using the Henderson–Hasselbalch equation:

pH = pKa + log [HCOONa]/[HCOOH] = $$\ce{3.75 + 0.60 = 4.35}$$

So my answer is that the pH of the buffer is $$\ce{4.35}$$

However, in our answer key, they found the new concentration of HCOOH and HCOONa by simply taking:

[HCOOH] = $$\ce{(0.05 *0.3)/0.08 = 0.1875M}$$

[HCOONa] = $$\ce{(0.03 *0.4)/0.08 = 0.15M}$$

and when using those concentrations in the Henderson–Hasselbalch equation they get the answer pH = $$\ce{3.65}$$

Question

Is the answer key correct? If we are adding a base, then the pH should increase to become more basic, not decrease as in the answer key. If the answer key is correct, why does the pH decrease?

• The answer key's method is certainly valid given that the system is buffered. I suspect that your calculations for final concentrations are incorrect. For one thing, you should be doing those calculations on concentrations, since the equilibrium expression all use concentrations, not amount of substance. – Zhe Mar 8 at 16:35
• @Zhe But shouldn't the pH increase and become more basic? – katara Mar 8 at 17:06
• @katara - No look at your calculation of the moles. There are 0.015 moles of formic acid and only 0.012 moles of sodium formate. Thus you would expect the solution to be more acidic than the pKa value. – MaxW Mar 8 at 17:54
• @katara - you shot yourself in the foot by going off on a tangent and thinking about NaOH. It is true that if you started with 0.027 moles of formic acid and 0.012 moles of NaOH then you would end up with 0.015 moles of formic acid and 0.012 moles of sodium formate, but why go "backwards" to solve the problem? – MaxW Mar 8 at 18:43