Question

50 ml h2o is added to 50 ml solution of ba(oh)2 of strength 0.01 m. The ph value of the resulting solution is

Answer 1

Answer:- pH value of the resulting solution is 12.

Solution:- using the dilution equation we will calculate the molarity of the diluted solution.

$M_1V_1=M_2V_2$

Given:-

$M_1$ = 0.01 M

$V_1$ = 50 mL

$V_2$ = 50 + 50 = 100 mL

$M_2$ = ?

Let's plug in the values and solve it for the molarity of diluted solution.

$0.01M(50mL)=M_2(100mL)$

$M_2=\frac{0.01M(50mL)}{100mL}$

$M_2=0.005M$

$Ba(OH)_2(aq)\rightarrow Ba^+^2(aq)+2OH^-(aq)$

From above equation, 1 mol of barium hydroxide gives 2 moles of hydroxide ions. So, the concentration of hydroxide ions would also be 2 times the concentration of diluted solution of barium hydroxide.

$[OH^-]=2(0.005M)$ = 0.01M

$pOH=-log[OH^-]$

$pOH=-log0.01$

pOH = 2

pH = 14 - pOH

pH = 14 - 2

pH = 12

So, the pH value of the resulting solution is 12.