A 0.6845 g sample of an unknown monoprotic acid requires 28.4 mL of 0.553 M NaOH for neutralization to a phenolphthalein end point. There are 7.35 mL of 0.914 M HCl used for back-titration. a. How many moles of OH¯ are used? Moles of OH moles OH" How many moles of H+ from HCl? Moles of H+ moles H* b. How many moles of H* are there in the solid acid? Moles H+ in solid acid = Moles OH in NAOH soln. – Moles Ht in HCl soln. Moles of H+ in the solid acid = moles H* in solid c. What is the molar mass of the unknown acid? с. Grams of acid Molar Mass = Moles of Ht ion furnished Molar Mass of the unknown acid |g/mol

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Chapter1: Chemical Foundations
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A 0.6845 g sample of an unknown monoprotic acid requires 28.4 mL of 0.553 M NaOH for neutralization to a
phenolphthalein end point. There are 7.35 mL of 0.914 M HCl used for back-titration.
a. How many moles of OH¯ are used?
Moles of OH
moles OH"
How many moles of H+ from HCl?
Moles of H+
moles H*
b. How many moles of H* are there in the solid acid?
Moles H+ in solid acid = Moles OH in NAOH soln. – Moles Ht in HCl soln.
Moles of H+ in the solid acid =
moles H* in solid
c. What is the molar mass of the unknown acid?
с.
Grams of acid
Molar Mass =
Moles of Ht ion furnished
Molar Mass of the unknown acid
|g/mol
Transcribed Image Text:A 0.6845 g sample of an unknown monoprotic acid requires 28.4 mL of 0.553 M NaOH for neutralization to a phenolphthalein end point. There are 7.35 mL of 0.914 M HCl used for back-titration. a. How many moles of OH¯ are used? Moles of OH moles OH" How many moles of H+ from HCl? Moles of H+ moles H* b. How many moles of H* are there in the solid acid? Moles H+ in solid acid = Moles OH in NAOH soln. – Moles Ht in HCl soln. Moles of H+ in the solid acid = moles H* in solid c. What is the molar mass of the unknown acid? с. Grams of acid Molar Mass = Moles of Ht ion furnished Molar Mass of the unknown acid |g/mol
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