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

Sagot :

MrRoyal

Answer:

[tex]M_b = \frac{0.10M * 9.50mL}{3.80mL}[/tex] ---- The setup

[tex]M_b = 0.25M[/tex] --- 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:

[tex]V_a = Final\ Reading - Initial\ Reading[/tex]

[tex]V_a = 25.00mL - 15.50mL[/tex]

[tex]V_a = 9.50mL[/tex]

Calculate the final volume of base (KOH) used:

[tex]V_b = Final\ Reading - Initial\ Reading[/tex]

[tex]V_b = 8.80mL - 5.00mL[/tex]

[tex]V_b = 3.80mL[/tex]

The numerical setup is calculated using::

[tex]M_a * V_a = M_b * V_b[/tex]

Where

[tex]V_a = 9.50mL[/tex]

[tex]V_b = 3.80mL[/tex]

[tex]M_a = 0.10M[/tex] --- the given molarity of HCL

So, we have:

[tex]M_a * V_a = M_b * V_b[/tex]

[tex]0.10M * 9.50mL = M_b * 3.80mL[/tex]

Make Mb the subject

[tex]M_b = \frac{0.10M * 9.50mL}{3.80mL}[/tex] ---- The correct numerical setup

The solution is then calculated as:

[tex]M_b = \frac{0.10M * 9.50mL}{3.80mL}[/tex]

[tex]M_b = \frac{0.10 * 9.50}{3.80}M[/tex]

[tex]M_b = \frac{0.95}{3.80}M[/tex]

[tex]M_b = 0.25M[/tex]

View image MrRoyal
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