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Calculate the Ph of the Resultant Mixtures: 10 Ml of 0.2m Ca(Oh)2 + 25 Ml of 0.1m Hcl - Chemistry

Calculate the pH of the resultant mixtures: 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl

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Solution

Moles of `H_3O^+` = `(25  xx 0.1)/1000` = .0025 mol

Moles of `OH^(-)` = `(10 xx 0.2 xx 2)/1000 = .0040 mol`

Thus, excess of `OH^(-)` = .0015 mol

`[OH^(-)] = .0015/(35xx 10^(-3)) "mol/L" = .0428`

`pOH = -log [OH]`

= 1.36

pH = 14 - 1.36

= 12.63 (not matched)

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APPEARS IN

NCERT Class 11 Chemistry Textbook
Chapter 7 Equilibrium
Q 66.1 | Page 230
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