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 .

Express the pH to two decimal places.

Answer 1

The pH of the solution is 12.31 .

Step-by-step explanation:

Initial molarity of barium hydroxide =
`M_1=6.5* 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=(6.5* 10^(-2)M* 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* 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 .