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

`M_b = (0.10M * 9.50mL)/(3.80mL)` ---- The setup

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

Step-by-step 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 = (0.10M * 9.50mL)/(3.80mL)` ---- The correct numerical setup

The solution is then calculated as:

`M_b = (0.10M * 9.50mL)/(3.80mL)`

`M_b = (0.10 * 9.50)/(3.80)M`

`M_b = (0.95)/(3.80)M`

`M_b = 0.25M`