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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]
