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Calculate the pH of each solution.(a) [OH?] = 1.9 × 10?7 M(b) [OH?] = 2.6 × 10?8 M(c)

Chapter 14, Problem 73P

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QUESTION:

Calculate the pH of each solution.

(a) \(\mathrm{[OH^-]=1.9 \times 10^{-7} ~M}\)

(b) \(\mathrm{[OH^-]=2.6 \times 10^{-8} ~M}\)

(c) \(\mathrm{[OH^-]=7.2 \times 10^{-11} ~M}\)

(d) \(\mathrm{[OH^-]=9.5 \times 10^{-2} ~M}\)

Equation Transcription:

Text Transcription:

[OH^-]=1.9 x 10^{-7} M

[OH^-]=2.6 x 10^{-8} M

[OH^-]=7.2 x 10^{-11} M

[OH^-]=9.5 x 10^{-2} M

Questions & Answers

QUESTION:

Calculate the pH of each solution.

(a) \(\mathrm{[OH^-]=1.9 \times 10^{-7} ~M}\)

(b) \(\mathrm{[OH^-]=2.6 \times 10^{-8} ~M}\)

(c) \(\mathrm{[OH^-]=7.2 \times 10^{-11} ~M}\)

(d) \(\mathrm{[OH^-]=9.5 \times 10^{-2} ~M}\)

Equation Transcription:

Text Transcription:

[OH^-]=1.9 x 10^{-7} M

[OH^-]=2.6 x 10^{-8} M

[OH^-]=7.2 x 10^{-11} M

[OH^-]=9.5 x 10^{-2} M

ANSWER:

Solution 73P :

Step 1:

Here, we have to pH of each solution :

pOH is the measure of concentration of hydroxide ion (OH-) in the solution. It is used to measure the alkalinity of a solution.

To calculate the pOH of a solution we should know the concentration of the hydroxide ion in moles per liter, i.e the molarity of the solution.

pOH is calculated using the expression:
                                        pOH =  - log [OH-]

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