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

When mixing, the reaction occurs

HCOOH + NaOH=HCOONa + H2O

number of moles of initial reagents:

n = c*V

n(HCOOH) = 0.225 mol

n (NaOH) = 0.0225 mol

after the reaction 0.0225 mol of sodium formate is formed and 0.2025 mol of methanoic acid remains

the total volume of the solution will be 180 ml (the water released in the reaction can be ignored)

C (acids) = n/V = 1.125 M, C (salts) = 0.125 M

pH of the buffer solution weak acid + salt:

pH = - lgKa-lg(C(K)/S (salt)) =-lg(1.78*10^-4)-lg(1.125/0.125)=3.74958-0.9542=2.7953