Question

Question 3 1 pts A student performs part C of the substitution lab. The student is using 0.50 Mbutyl 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 HCI, what would be the [OH-] from Data Set C of the student data (shown below? Report answers with 4 places past the decimal...for example 0.000158, would be 0.0002 Data Set Time (sec) Initial buret final buret ml HCI (OH) 1/(OH) A O seconds 0.15 mL 10.14 mL 615 seconds 10.14 mL 15.21 mL 1290 seconds 15.21 mL 17.61 mL 1915 seconds 17.61 ml 18.99 mL

Answer 1

Data Set	mL HCl	[OH-]	1/[OH-]
A	9.99 mL	0.0749 M	13.3467 M-1
B	5.07 mL	0.0380 M	26.2985 M-1
C	2.40 mL	0.0180 M	55.5556 M-1
D	1.38 mL	0.0104 M	96.6184 M-1

Explanation

For data set A,

final buret reading = 10.14 mL

initial buret reading = 0.15 mL

volume HCl used = (final buret reading) - (initial buret reading)

volume HCl used = (10.14 mL) - (0.15 mL)

volume HCl used = 9.99 mL

moles HCl used = (molarity HCl) * (volume HCl used)

moles HCl used = (0.15 M) * (9.99 mL)

moles HCl used = 1.4985 mmol

moles OH^- = moles HCl used

moles OH^- = 1.4985 mmol

[OH^-] = (moles OH^-) / (volume of solution)

[OH^-] = (1.4985 mmol) / (20 mL)

[OH^-] = 0.0749 M

1/[OH^-] = 1/(0.0749 M)

1/[OH^-] = 13.3467 M^-1