Question

Suppose you carry out a titration involving 3.00 molar HCl and an unknown concentration of KOH. To bring the reaction to its end point, you add 35.3 milliliters of HCl to 105.0 milliliters of KOH. What is the concentration of the KOH solution?

Answer 1

Answer: The concentration of the KOH solution is 1.01 M

Explanation:

According to neutralization law:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1$ = basicity of $HCl$ = 1

$n_2$ = acidity of KOH = 1

$M_1$ = concentration of HCl = 3.00 M

$M_2$ = concentration of KOH = ?

$V_1$ = volume of HCl = 35.3 ml

$V_2$ = volume of KOH = 105.0 ml

Putting the values we get:

$1\times 3.00\times 35.3=1\times M_2\times 105.0$

$M_2=1.01$

Thus the concentration of the KOH solution is 1.01 M