Question

A solution is prepared by diluting 43.5 mL of 6.5×10?2M Ba(OH)2 to a volume of 270.5 mL .

Answer 1

Answer:

The pH of the solution is 12.31 .

Explanation:

Initial molarity of barium hydroxide =$M_1=6.5\times 10^{-2}M$

Initial volume of barium hydroxide =$V_1=43.5 mL$

Final molarity of barium hydroxide =$M_2$

Final volume of barium hydroxide =$V_2=270.5 mL$

$M_1V_1=M_2V_2$

$M_2=\frac{6.5\times 10^{-2}M\times 43.5 mL}{270.5 mL}$

$M_2=0.0104 M$

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

1 mol of barium gives 2 mol of hydroxide ions.

Then 0.0104 M of barium hydroxide will give:

$2\times 0.0104 M=0.0208 M$ of hydroxide ions

$[OH^-]=0.0208 M$

$pH=14-pOH=14-(-\log[OH^-])$

$pH=14-(-\log[0.0208 M])=12.31$

The pH of the solution is 12.31 .