See image below. Can you please show me how to solve this? This problem had parts a through d... **Exercise: Calculating pH of Solutions**Calculate the pH of the following solutions:a) \(1.00 \times 10^{-10} \, \text{M} \, \text{NaOH}\)b) \(1.00 \times 10^{-2} \, \text{M} \, \text{NaOH}\)c) \(1.00 \times 10^{-1} \, \text{M} \, \text{HC}_2\text{H}_3\text{O}_2\)d) \(1.00 \times 10^{-1} \, \text{M} \, \text{NaC}_2\text{H}_3\text{O}_2\)**Instructions:**To solve each part, consider the properties of the given compound—whether it's a strong base or weak acid/base—and use appropriate formulas to determine the pH. Use the dissociation principles and the concept of equilibrium constants when necessary.

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Introduction to Chemical Engineering Thermodynamics

8th Edition

ISBN:9781259696527

Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Chapter1: Introduction

Section: Chapter Questions

Problem 1.1P

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See image below. Can you please show me how to solve this? This problem had parts a through d...

**Exercise: Calculating pH of Solutions**

Calculate the pH of the following solutions:

a) \(1.00 \times 10^{-10} \, \text{M} \, \text{NaOH}\)

b) \(1.00 \times 10^{-2} \, \text{M} \, \text{NaOH}\)

c) \(1.00 \times 10^{-1} \, \text{M} \, \text{HC}_2\text{H}_3\text{O}_2\)

d) \(1.00 \times 10^{-1} \, \text{M} \, \text{NaC}_2\text{H}_3\text{O}_2\)

**Instructions:**

To solve each part, consider the properties of the given compound—whether it's a strong base or weak acid/base—and use appropriate formulas to determine the pH. Use the dissociation principles and the concept of equilibrium constants when necessary.

Expert Solution

Step 1

(i) 1.00×10^-10 M NaOH:

When we calculate pH, it is considered that the solution has NaOH as a given concentration in H₂O.

As when the acid or base is less than 10^-7 then, the solute can be considered to be very dilute. the OH^- ion is contributed from H₂O. Here the [] denotes the concentration.

Hence, [H⁺] = [OH^-] = 10^-7

pH = -log[H⁺] = 7

(ii) 1.00×10^-2 M NaOH:

pk_wfor water is 14.

pOH = -log[OH^-] = 2

pH = pk_w - pOH = 14-2 = 12

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