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Karnataka Board PUCPUC Science Class 11

Calculate the Ph of the Following Solutions: 0.3 G of Ca(Oh)2 Dissolved in Water to Give 500 Ml of Solution.

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Question

Calculate the pH of the following solutions:

0.3 g of Ca(OH)dissolved in water to give 500 mL of solution.

Numerical
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Solution

Strength of Ca(OH)2 = `0.3/500 xx 1000`
= 0.6 gL-1

Molar mass of Ca(OH)= 40 + 32 + 2 = 74

Molarity of Ca(OH)= `0.6/74` = 8.11 x 10-3M

1 mole of Ca(OH)= `0.6/74` = 8.11 x 10-3 M

1 mole of Ca(OH)on dissociation produces
two moles of OH-1 ions.

`therefore ["OH"^-] = 2xx8.11xx10^(-3)"M"`

= `16.22 x 10^(-3)"M"`

But `["H"^+]["OH"^-] = "k"_"w" = 1 xx 10^(-14)`

`therefore ["H"^+] = (1 xx 10^(-14))/["OH"^-] = (1 x 10^(-14))/(16.22 xx 10^(-3))`

= `6.16 xx 10^(-3)`M

Now pH = `- log["H"^+]`

= `- log[6.16 xx 10^(-13)]`

= `- [log 6.16 + log^(-13)]`

= `- [0.7896 - 13]`

`= 12.2104`

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Chapter 6: Equilibrium - EXERCISES [Page 237]

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NCERT Chemistry Part 1 and 2 [English] Class 11
Chapter 6 Equilibrium
EXERCISES | Q 7.49 - b) | Page 237

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