Question

150.00 mL of 1.50 mol/L methanoic acid is being titrated with 0.750 mol/L sodium hydroxide. What is the pH after 30.00 mL of base has been added?

(Ka=1.78 x 10−4)

Answer 1

Step I: Write balanced equation for the reaction,

HCOOH_(aq) + NaOH_(aq) $\to$ HCOONa_(aq) + H₂O_(l)

Mole ratio is 1:1

Initial Moles of HCOOH are calculated as;

1.5 moles are in 1000mL

? moles will be in 150mL

$=\dfrac{1.5mol x150mL}{1000mL}$

$=0.225 moles$

Moles of NaOH added is calculated as;

0.75 moles are in 1000mL

? moles will be in 30mL

$=\dfrac{0.75mol x 30mL}{1000mL}$

$=0.0225moles$

Therefore moles of HCOOH neutralized are = 0.0225 moles

And Moles of HCOOH remaining will be;

0.225-0.0225 = 0.02025 moles.

Step II: Concentration of acid and salt in the mixture

Total volume of solution becomes

150+30 = 180mL = 180/1000L =0.18L

Concentration of HCOOH remaining in the solution is calculated as;

$Concentration=\dfrac{number of moles}{Volume of solution in L}$

$Concentration = \dfrac{0.2025mol}{0.18L}$

= 1.125M

And Concentration of the salt (CHOONa) is;

$Concentration = \dfrac{0.0225mol}{0.18L}$

= 0.125M

Step III: Calculating pH

$pH=pKa + log\dfrac{[HCOONa]}{[HCOOH]}$

$pKa = -logKa = -log(1.78x10^-4) =3.75$

$\therefore pH = 3.75 + log\dfrac{0.125}{1.125}$

$pH = 3.75 + (-0.95) = 2.79$

pH of the solution is approximately 3