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

(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 = 0.0428M pOH = −log0.0428 = 1.368 pH = 14 − pOH = 14 − 1.368 = 12.635 (b) Total number of moles present in 10 mL of 0.01 M H2? SO4? is (10*0.01) / 1000? = 0.0001 mol. Total number of moles present in 10 mL of 0.01 M Ca (OH)2? = (10*0.01) / 1000? = 0.0001 mole. Ca (OH)2? + H2? SO4? → CaSO4? + 2H2? O 0.0001 mole of Ca (OH)2? will react completely with 0.0001 mole of H2? SO4? Hence, the resulting solution is neutral with pH 7.0. (c) Total number of moles in 10 mL of 0.1 M H2? SO4? = (10*0.1) / 1000? = 0.001 mole. Total number of moles present in 10 mL of 0.1 M KOH = (10*0.1? ) / 1000 = 0.001 mole. 2KOH + H2? SO4? → K2? SO4? + 2H2? O 2 moles of KOH reacts with 1 mole of H2? SO4. 0.001 mole of KOH will react with 0.0005 mole of H2? SO4? Number of moles of H2? SO4 left = 0.001−0.0005 = 0.005M Volume of solution is 10+10 = 20mL Molarity of the solution is (0.0005*1000? ) / 20 = 0.025M. [H+] = 2 * 2.5 * 10−2 = 0.05M pH = −log [H+] = −log0.05 = 1.3.

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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Vishal Baghel | Contributor-Level 10

9 months ago

(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 H