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Calculate the pH of each solution.(a) [OH?] = 2.8 × 10?11 M(b) [OH?] = 9.6 × 10?3 M(c)

Chapter 14, Problem 74P

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QUESTION:

Calculate the pH of each solution.

(a) \(\mathrm{[OH^-]=2.8 \times 10^{-11} ~M}\)

(b) \(\mathrm{[OH^-]=9.6 \times 10^{-3} ~M}\)

(c) \(\mathrm{[OH^-]=3.8 \times 10^{-12} ~M}\)

(d) \(\mathrm{[OH^-]=6.4 \times 10^{-4} ~M}\)

Equation Transcription:

Text Transcription:

[OH^-]=2.8 x 10^{-11} M

[OH^-]=9.6 x 10^{-3} M

[OH^-]=3.8 x 10^{-12} M

[OH^-]=6.4 x 10^{-4} M

Questions & Answers

QUESTION:

Calculate the pH of each solution.

(a) \(\mathrm{[OH^-]=2.8 \times 10^{-11} ~M}\)

(b) \(\mathrm{[OH^-]=9.6 \times 10^{-3} ~M}\)

(c) \(\mathrm{[OH^-]=3.8 \times 10^{-12} ~M}\)

(d) \(\mathrm{[OH^-]=6.4 \times 10^{-4} ~M}\)

Equation Transcription:

Text Transcription:

[OH^-]=2.8 x 10^{-11} M

[OH^-]=9.6 x 10^{-3} M

[OH^-]=3.8 x 10^{-12} M

[OH^-]=6.4 x 10^{-4} M

ANSWER:

Solution 74P :

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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