7.66. Calculate the pH of the resultant mixtures:
a) 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl
b) 10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2
c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH
7.66. Calculate the pH of the resultant mixtures:
a) 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl
b) 10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2
c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH
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1 Answer
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(a) Total number of moles present in 10 mL of 0.2 M calcium hydroxide are (10×0.2) / 1000? = 0.002 moles.
Total number of moles present in 25 mL of 0.1 M HCl are (25×2) / 1000? = 0.0025 moles.Ca (OH)2? + 2HCl → CaCl2? + 2H2? O
1 mole of calcium hydroxide reacts with 2 moles of HCl.
0.0025 moles of HCl will react with 0.00125 moles of calcium hydroxide.
Total number of moles of calcium hydroxide unreacted are 0.002−0.00125 = 0.00075 moles.
Total volume of the solution is 10 + 25 = 35 mL.
The molarity of the solution is (0.00075×1000) / 35? = 0.0214M
[OH−] = 2 × 0.0214...more
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