Question

Determine the pOH of the solution that is produced by mixing 9.48 mL of 7.76×10-2 M Ca(OH)2 with 86.6 mL of 6.12×10-3 M NaH. (With steps please :)

Answer 1

CaH2 + 2 NaOH = Ca(OH)2 + 2 NaH

Total number of moles present in 9.48 mL of 7.76×10-2 M calcium hydroxide are

$\\ \frac{9.48 \times 7.76}{100000} = \frac{73.5648}{100000} \\ \\ = 0.000735648 \: moles$

Total number of moles present in 86.6 mL of 6.12×10-3 M NaH are

$\\ \frac{86.6 \times 6.12}{1000000} = \frac{529.992}{1000000} \\ \\ = 0.000529992 \: moles$

1 mole of calcium hydroxide reacts with 2 moles of NaH.

So,

0.00529992*2 = 0.001059984 moles

0.001059984 moles of NaH will react with 0.0007356 moles of calium hydroxide.

Total number of moles of calcium hydroxide unreacted are 0.001059984 − 0.000735648 = 0.000324336 moles.

Total volume of the solution is 9.48 + 86.6 = 96.08 mL.

The molarity of the solution is

$\frac{0.000324336 \times 1000}{96.08} \\ = \frac{0.324336}{96.08} \\ = \: 0.00337568693 \: M</p><p></p><p>$

[OH-] = 2*0.0033756 = 0.00675137386 M

pOH = - log(0.00675137386) = 2.17

&

pH = 14 - 2.17 = 11.83