Question 15.6: A 25.00-mL sample of H2SO4 solution required 14.26 mL of 0.2...

READ     Knowns     14.26 mL 0.2240 M NaOH solution

25.00 mL H_2SO_4 solution

Solving for:  sulfuric acid molarity

PLAN      The equation for the reaction is

2 NaOH(aq) + H_2SO_4 → Na_2SO_4(aq) + 2 H_2O(l)

Determine the moles NaOH in solution

SETUP            (0.01426  \cancel{L})(\frac{0.2240   mol   NaOH}{1   \cancel{L}} )=0.003194 mol NaOH

Since the mole ratio of acid to base is  \frac{1  mol  H_2SO_4 }{2   mol  NaOH},

(0.0031940   mol   NaOH) (\frac{1  mol  H_2SO_4 }{2   mol   NaOH} )=0.001597 mol H_2SO_4

CALCULATE     M=\frac{mol}{L} =\frac{0.001597   mol H_2SO_4 }{0.02500   L} =0.06388 M H2SO4