Question

How many milliliters of 0.048 M NaOH are required to reach the second equivalence point when titrating 28.0 mL of 0.018 M H2SO4?

A. 15.6

B. 23.8

C. 10.5

D. 21.0

E. 22.7

Answer 1

(0.018 mol/L) (0.0280 L) = 0.000504 mol of H2SO4

H2SO4 + 2NaOH ---> Na2SO4 + 2H2O

H2SO4 and NaOH react in a 1:2 molar ratio

1 is to 2 as 0.000504 is to x

x = 0.001008 mol of NaOH required

(0.048 mol/L) (x) = 0.001008 mol

x = 0.021 L = 21.0 mL

So the correct option is D. 21.0 mL

Or

2 NaOH + H2SO4 ------> Na2SO4 + 2 HOH (2 H2O)

So we need 2 mole NaOH/Mole H2SO4

Mole sulfuric acid = 0.028 L x 0.018 mole/ L = 0.000504
Mole NaOH = 0.00504 mole H2SO4 x 2mol NaOH/mol H2SO4 = 0.001008 mol NaOH

mL NaOH = 1 L NaOH/0.048 mol NaOH x .001008 mol NaOH = 0.021 liter = 21 mL