Can someone help me to solve this Chemistry problem and explain how?

Question

What volume of 0.0500M sodium hydroxide should be added to 250.mL of 0.100M HCOOH to obtain a solution with a pH of 4.50? (Ka(HCOOH)=1.7X10-4)

Possible answers:
A. 540mL
B. 420mL
C. 80mL
D. 340mL
E. 500mL

Answer 1

Step 1. Look at the nature of the reactants.
You have a reaction of a stong base and weak acid.

Step 2. Look at the desired pH. You want acidic pH. That means that the acid will be in excess. Also since we said in step 1 that we have a weak acid and a strong base we will be forming a buffer system. In order to get the most accurate solution we would need to set-up an ICE table. However we can get an approximate solution which most of the times will be the same as the exact using the Henderson-Hasselbalch equations (and assumptions).

pH=pKa+ log( [conj.base]/[acid] )

Step 3. Write the equation in order to determine the stoichiometry
HCOOH + NaOH- -> HCOONa +H2O
So the stoichiometry is 1 HCOOH : 1NaOH: 1 HCOONa

Step 4. Find the number of moles of each compound
If we used Vb L of NaOH then we used
mole NaOH= MbVb
from step 3, we know that we produced
mole HCOONa=mole NaOH, thus mole HCOONa= MbVb
we had mole HCOOH=MaVa
so after the reaction remained mole HCOOH= MaVa-MbVb

Step 5. Convert moles to concentrations. The volume of the solution becomes V=Va+Vb
[conj.base]= [HCOO-] =[HCOONa] = MbVb/(Va+Vb)
[acid]= [HCOOH]= (MaVa-MbVb)/(Va+Vb)

Step 6 Divide the concentrations and then substitute in the equation.
[conj.base]/[acid]= MbVb / (MaVa-MbVb)
so
pH= pKa+log (MbVb / (MaVa-MbVb)) =>
MbVb / (MaVa-MbVb)=10^(pH-pKa) =>
0.05*Vb/ (0.1*0.25-0.05Vb)= 10^(4.5-(-log(1.7*10^-4)) )=5.38 =>
0.05 Vb= 0.1345-0.269Vb =>
Vb=0.1345/0.319 =0.422 L =422 mL

So the answer is B (There is a slight deviation probably because of the rounding of the 10^(pH-pKa))