Solution Found!
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(choose chapter or problem)
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-]