(a) By titration, 15.0 mL of 0.1008 M sodium hydroxide is needed
...

Question

(a) By titration, 15.0 mL of 0.1008 M sodium hydroxide is needed
to neutralize a 0.2053-g sample of a weak acid. What is the molar
mass of the acid if it is monoprotic?

Accepted Answer

Hi everyone here we have a question asking us to determine the molar mass of a 0.1093 g week monoprix acid that is neutralized by titrate of 9.86 mL of 0.1787 moller potassium hydroxide at the equivalence point of titrate. A shin. The moles, a potassium hydroxide added equals the moles of hydrogen present. So our polarity of potassium hydroxide times. The leaders of potassium hydroxide equals the grandes of acid, divided by the molar mass of acid. So we have middle liters. We have 9.86 ml and we need to change that to leaders. So we're going to multiply that buy one liter Over to the 3rd ml. And our middle leaders are going to cancel out, giving us 0. 986 L. So the molar mass other acid is going to equal zero 1093g of acid divided by 0.1787 moller potassium hydroxide times 0. leaders of potassium hydroxide And that equals .03 g per mole. And that is our final answer. Thank you for watching. Bye.

Watch next