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Get Full Access to Introductory Chemistry - 5 Edition - Chapter 14 - Problem 78p
Get Full Access to Introductory Chemistry - 5 Edition - Chapter 14 - Problem 78p

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# Calculate the pH of each solution:(a) 1.34 × 10?3 M HClO4(b) 0.0211 M NaOH(c) 0.0109 M

ISBN: 9780321910295 34

## Solution for problem 78P Chapter 14

Introductory Chemistry | 5th Edition

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Problem 78P

Calculate the pH of each solution:

(a) 1.34 × 10−3 M HClO4

(b) 0.0211 M NaOH

(c) 0.0109 M HBr

(d) 7.02 × 10−5 M Ba(OH)2

Step-by-Step Solution:
Step 1 of 3

Solution 78P:

Here, we are going to calculate the pH of each solution.

pH is defined as the negative log of hydrogen or hydronium ion concentration.

pH = - log[H+ ]

(a) 1.34 × 10−3 M HClO4

We know,

pH =  - log[H+]

= - log (1.34 × 10−3 )

= -log 1.34 - log 10−3

= -0.127 +3 = 2.87

Thus, pH of 1.34 × 10−3 M HClO4 is 2.87.

(b) 0.0211 M NaOH

We know,

pOH =  - log[OH-]

= - log (0.0211 )

= -(-1.676) =1.676

Thus, we know

pH+pOH =14

pH = 14-pOH

=14 - 1.676 =12.32

Thus, pH of 0.0211 M NaOH is 12.32.

(c) 0.0109 M HBr

We know,

pH =  - log[H+]

= - log (0.0109 )

= -(-1.96)

= 1.96

Thus, pH of  0.0109 M HBr is 1.96.

(d) 7.02 × 10−5 M Ba(OH)2

We know,

[OH-] = 2 x  7.02 × 10−5  = 14.04 x 10−5 M

pOH =  - log[OH-]

= - log (14.04 × 10−5 )

= -log 14.04-log10−5

=-1.147+5

= 3.85

Thus, we know

pH+pOH =14

pH = 14-3.85

=10.147

Thus, pH of 7.02 × 10−5 M Ba(OH)2 is 10.15

Step 2 of 3

Step 3 of 3

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