What is the concentration of a $K O H (a q)$ $KOH (aq)$ solution if 12.8 mL of this solution is required to react with 25.0 mL of .110 mol/L $H_{2} S O_{4} (a q)$ $H_2SO_4(aq)$ ?

Question

14479 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-17T10:35:24

We assume stoichiometric quantities of potassium hydroxide and sulfuric acid.

$2 K O H (a q) + H_{2} S O_{4} \to K_{2} S O_{4} (a q) + 2 H_{2} O (l)$ $2KOH(aq) + H_2SO_4 rarr K_2SO_4(aq) + 2H_2O(l)$

The reaction represents the neutralization of sulfuric acid by potassium hydroxide.

Moles of sulfuric acid: $25.0xx10^-3cancelLxx0.110*mol*cancel(L^-1)$ $=$ $??$ $mol$

From the equation above, which shows that the stoichiometry of hydroxide to sulfuric acid is $2:1$ , and the volume of stoichiometric acid:

$[KOH]=(25.0xx10^-3cancelLxx0.110*mol*cancel(L^-1)xx2)/(12.8xx10^-3L)$ $=$ $??mol*L^-1$ .

We get an answer in $mol*L^-1$ , as required.

What is the concentration of a KOH (aq)KOH(aq)KOH (aq) solution if 12.8 mL of this solution is required to react with 25.0 mL of .110 mol/L H_2SO_4(aq)H2SO4(aq)H_2SO_4(aq)?