The following questions refer to the titration of 25.00 mL of0.100 M HCl with 0.
ID: 687902 • Letter: T
Question
The following questions refer to the titration of 25.00 mL of0.100 M HCl with 0.100 M Ba(OH)2.The reaction proceedsaccording to the following stoichiometry. 2HCl + Ba(OH)2 ---> BaCl2 +2H2O a) What will the pH of the titrated solution be after theaddition of 10.00 mL of Ba(OH)2? b) What will be the pH of the titrated solution at theequivalence point? c) What will the pH of the titrated solution be after tehaddition of 15.00 mL of Ba(OH)2? The following questions refer to the titration of 25.00 mL of0.100 M HCl with 0.100 M Ba(OH)2.The reaction proceedsaccording to the following stoichiometry. 2HCl + Ba(OH)2 ---> BaCl2 +2H2O a) What will the pH of the titrated solution be after theaddition of 10.00 mL of Ba(OH)2? b) What will be the pH of the titrated solution at theequivalence point? c) What will the pH of the titrated solution be after tehaddition of 15.00 mL of Ba(OH)2?Explanation / Answer
Balanced chemical reaction:
2HCl (aq) + Ba(OH)2 (aq) ---> BaCl2 (aq) + 2H2O (l)
Beforerxn 0.1 M * 0.025 L 0.1 M * 0.01 L
= 0.0025 moles 0.001
Afterrxn 0.0015moles 0 0.001
[H+] = 0.0015 moles / 0.025 L + 0.01 L
= 0.0428 M
pH = 1.36
b) Being astrong acid strong base titration, pH at the equivalence point is7.
c) 2HCl (aq) + Ba(OH)2 (aq) ---> BaCl2 (aq) + 2H2O (l)
Beforerxn 0.1 M * 0.025 L 0.1 M * 0.015 L
= 0.0025 moles 0.0015
Afterrxn 0.001moles 0 0.001
[H+] = 0.001 moles / 0.025 L + 0.015 L
= 0.025 M
pH = 1.6
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