Answered

Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Explore in-depth answers to your questions from a knowledgeable community of experts across different fields. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

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
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.