1.

Determine the pH of the solution that results from the addition of 20.00 mL of 0.01 M Ca(OH)_(2) to 30.00 mL of 0.01 MHCl

Answer»

11.3
10.53
`2.70`
8.35

Solution :Millimoles of `H^(+) = 30 XX 0.01 = 0.3`
Millimoles of `OH^(-) = 20 xx 0.01 xx 2 = 0.4`
Remaining millimoles of `OH^(-) = 0.4 - 0.3 = 0.1`
`[OH^(-)] = (0.1)/(50)` or `2 xx 10^(-3)`
So, `pOH = 2.6999 rArr PH = 14 - 2.6999 = 11.30`.


Discussion

No Comment Found

Related InterviewSolutions