Suppose that 42.00 mL of 0.150 M NaOH solution are required to neutralize 50.00 mL of H2SO4 solution. What is the molarity of the acid solution?

Question

Suppose that 42.00 mL of 0.150 M NaOH solution are required to neutralize 50.00 mL of [latex]H_2SO_4[/latex] solution. What is the molarity of the acid solution?

Accepted Answer

READ Knowns 42.00 mL 0.150 M NaOH solution
50.00 mL [latex]H_2SO_4[/latex] solution

Solving for: acid molarity
PLAN The equation for the reaction is

2 NaOH(aq) + [latex]H_2SO_4 [/latex](aq) → [latex]Na_2SO_4 [/latex](aq) + 2 [latex]H_2O [/latex](l)

Determine the moles NaOH in the solution.

SETUP (0.04200 [latex]\cancel{L} [/latex]) [latex](\frac{0.150 mol NaOH}{1 \cancel{L}} ) [/latex]= 0.00630 mol NaOH

Since the acid and base react in a 1:2 ratio (from the equation for the
reaction) 2 mol base = 1 mol acid.

(0.00630 [latex]\cancel{mol NaOH}[/latex]) [latex](\frac{1 mol H_2SO_4}{2 \cancel{mol NaOH}} ) [/latex]= 0.00315 mol [latex]H_2SO_4 [/latex]

CALCULATE [latex] M=\frac{mol}{L} =\frac{0.00315 mol H_2SO_4}{0.05000 L} =0.0630 M H_2SO_4 [/latex]

Question 15.5: Suppose that 42.00 mL of 0.150 M NaOH solution are required ...

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