Question

In a titration, 15.65 milliliters of a KOH * (aq) ) solution exactly neutralized 10.00 milliliters of a 1.22 M HCl * (aq) solution . In the space below , show a correct numerical setup for calculating the molarity of the KOH * (aq) solution .

Answer 1

Answer:

The molarity of the KOH solution is 0.780 M.

Explanation:

$KOH(aq)+HCl(aq)\rightarrow KCl(aq)+H_2O(l)$

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is KOH

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is HCl.

We are given:

$n_1=1\\M_1=?\\V_1=15.65 mL\\n_2=1\\M_2=1.22 M\\V_2=10.00 mL$

Putting values in above equation, we get:

$1\times M_1\times 15.65 mL=1\times 1.22 M\times 10.00 mL$

$M_1=\frac{1\times 1.22 M\times 10.00 mL}{1\times 15.65 mL}$

$M_1=0.7795 M \approx 0.780 M$

The molarity of the KOH solution is 0.780 M.

Answer 2

Answer: HCL(aq) + KOH(Aq) —> H2O + KCL

Explanation:

KOH is a base and HCL is an acid. Acids and bases mixed result in salt and water. So the products are H2O and KCL.