In a titration experiment, 12.5 mL of 0.500 M $H_2SO_4$ neutralizes 50.0 mL of $NaOH$ . What is the concentration of the $NaOH$ solution?

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Answer 1 · 2017-08-04T03:14:34

0.25 M

$0.500 (mol)/L H_2SO_4 * 0.0125 L = 0.00625 mol H_2SO_4$

Write a balanced equation for the reaction:

$H_2SO_4 + 2NaOH -> 2H_2O + Na_2SO_4$

1 mol $H_2SO4$ neutralizes 2 mol $NaOH$ , so:

$0.00625*2 = 0.0125 mol NaOH$

$0.0125 mol NaOH / 0.05 L= 0.25 M NaOH$

In a titration experiment, 12.5 mL of 0.500 M H_2SO_4H_2SO_4 neutralizes 50.0 mL of NaOHNaOH. What is the concentration of the NaOHNaOH solution?