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Question
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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Solution
Strength of Ca(OH)2 = `0.3/500 xx 1000`
= 0.6 gL-1
Molar mass of Ca(OH)2 = 40 + 32 + 2 = 74
Molarity of Ca(OH)2 = `0.6/74` = 8.11 x 10-3M
1 mole of Ca(OH)2 = `0.6/74` = 8.11 x 10-3 M
1 mole of Ca(OH)2 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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