Question

I have the answer, im just not sure how to do the problem.

Question 3 0/1 pts A student performs part of the substitution lab. The student is using 0.50 M butyl bromine in ethanol, 0.50 M potassium hydroxide in ethanol and 0.15 M HCl for the reactions. If the student took a 20 mL sample of the solution and titrated it with HCl, what would be the (OH) from Data Set A of the student data (shown below)? Report answers with 4 places past the decimal...for example 0.000158, would be 0.0002 Duta Set Time (sec) Initial buret final buret јон 1/(OH) O seconds 0.15 ml 10.14 mL 615 seconds 10.14 mL 15.21 mL 1290 seconds 15.21 mi 17.61 ml 1915 seconds 1761 ml 18.99 L You Answered 0.0006 Correct Answers Between 0.07 and 0.08

I think they are mixed

Answer 1

Student took 20.0 ml of the Solution.

Titrated with : 0.15 M HCl

Volume of HCl used = (10.14 - 0.15 ) ml = 9.99 ml (from dataset A)

Reaction occured :

H⁺ + OH^- -------> H₂O (other substances do not consume HCl0

Since 1mol of HCl reacts with one mole of ( OH^-).

Moles of HCl reacted =  9.99 ml * 0.15 (mol/L) / 1000 ml/ L = 1.4985*10^-3 moles

since , 1mol of HCl = 1 mole of ( OH^-).

So, moles of ( OH^-). reacted =  1.4985*10^-3 moles

This in 20 ml solution,

So, molarity of ( OH^-) = (1.4985*10^-3 moles * 1000 ml/L) / 20 ml = 0.0749 mol / L

or  ( OH^-) = 0.0749 M