Question

20. Show a correct numerical setup for calculating the molarity of the KOH * (aq) solution. Then state the calculated value of the molarity .

Answer 1

Answer:

$M_b = \frac{0.10M * 9.50mL}{3.80mL}$ ---- The setup

$M_b = 0.25M$ --- The molarity of KOH

Explanation:

Given

I will answer the question with the attached titration data

Required

The set and the value of the molarity of KOH

First, calculate the volume of acid (HCL) used:

$V_a = Final\ Reading - Initial\ Reading$

$V_a = 25.00mL - 15.50mL$

$V_a = 9.50mL$

Calculate the final volume of base (KOH) used:

$V_b = Final\ Reading - Initial\ Reading$

$V_b = 8.80mL - 5.00mL$

$V_b = 3.80mL$

The numerical setup is calculated using::

$M_a * V_a = M_b * V_b$

Where

$V_a = 9.50mL$

$V_b = 3.80mL$

$M_a = 0.10M$ --- the given molarity of HCL

So, we have:

$M_a * V_a = M_b * V_b$

$0.10M * 9.50mL = M_b * 3.80mL$

Make Mb the subject

$M_b = \frac{0.10M * 9.50mL}{3.80mL}$ ---- The correct numerical setup

The solution is then calculated as:

$M_b = \frac{0.10M * 9.50mL}{3.80mL}$

$M_b = \frac{0.10 * 9.50}{3.80}M$

$M_b = \frac{0.95}{3.80}M$

$M_b = 0.25M$