1. 56.3 ml of 1.05 M hydrochloric acid is added to 25.1 ml of barium hydroxide,a
ID: 1033972 • Letter: 1
Question
1. 56.3 ml of 1.05 M hydrochloric acid is added to 25.1 ml of barium hydroxide,and the resulting solution is found to be acidic. 15.5 ml of 2.76 M sodium hydroxide is required to reach neutrality. What is the polarity of the original barium hydroxide solution?
2. 45.4 ml of 0.498 M nitric acid is added to 25.3 ml of calcium hydroxide, and the resulting solution is found to be acidic. 25.1 ml of 0.236 M barium hydroxide is required to reach neutrality. what is the polarity of the original calcium hydroxide?
Explanation / Answer
1. Ba(OH)2 + 2HCl --> BaCl2 + 2H2O
HCl + NaOH --> NaCl + H2O
Total moles of HCl added = 1.05 M x 56.3 ml = 59.115 mmol
excess moles HCl = 2.76 M x 15.5 ml = 42.78 mmol
moles HCl consumed by Ba(OH)2 = 59.115 - 42.78 = 16.335 mmol
moles Ba(OH)2 neutralized = 16.335 mmol/2 = 8.1675 mmol
molarity of Ba(OH)2 solution = 8.1675 mmol/25.1 ml = 0.3254 M
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2. Ca(OH)2 + 2HNO3 --> Ca(NO3)2 + 2H2O
Ba(OH)2 + 2HNO3 --> Ba(NO3)2 + 2H2O
Total moles of HNO3 added = 0.498 M x 45.4 ml = 22.61 mmol
excess moles HNO3 = 0.236 M x 25.1 ml = 5.924 mmol
moles HNO3 consumed by Ca(OH)2 = 22.61 - 5.924 = 16.686 mmol
moles Ca(OH)2 neutralized = 16.686 mmol/2 = 8.343 mmol
molarity of Ca(OH)2 solution = 8.343 mmol/25.3 ml = 0.330 M
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