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Chapter 10: Q. 10.87 (page 359)

Determine the pH for the follow ing solutions: $(10, 6)$
a. $[OH] = 1.0 脳 10^{- 7} M$
b. $[H_{2} O^{'}] = 4.2 脳 10^{- 3} M$
c. $[H_{2} O^{+}] = 0.0001 M$
d. $[OH] = 8.5 脳 10^{- 9} M$

Short Answer

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Step by step solution

01

Step 1 Introduction

The $pH$ scale relates to the logarithmic scale. $[H_{3} O^{+}]$ of aqueous solutions.

Mathematically expression of $pH$ is the negative logarithmic of the $[H_{3} O^{+}]$

$K w = [H_{3} O^{+}] [{OH}^{-}]$

$pH = - \log [H_{3} O^{+}]$

$pOH = - \log [{OH}^{-}]$

Acidic solution $= pH < 7.0 [H_{3} O^{+}] > 1.0 脳 10^{- 7} M$

Neutral solution $= pH = 7.0 [H_{3} O^{+}] = 1.0 脳 10^{- 7} M$

Basic solution $= pH > 7.0 [H_{3} O^{+}] < 1.0 脳 10^{- 7} M$

02

Step 2 Explantion part (a)

Given acid concentration of the solution is $1.0 脳 10^{- 11} M$ . Calculate the $pH$ of the solution by using the following formula:

$p H = - \log [H_{3} O^{+}]$

$= - \log [1.0 脳 10^{- 11}]$

$= 11 - \log (1.0)$

$= 11$

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Most popular questions from this chapter

When does a reversible reaction reach equilibrium?

Identify each of the solutions in problem 10.88 as acidic, basic, or neutral.

Why does a neutral solution have a pH of $7.0$ ?

Balance each of the following neutralization equations:

a. ${HNO}_{3} (a q) + Ba (OH)_{2} (s) 鉄� H_{2} O (l) + Ba {({NO}_{3})}_{2} (a q)$

b.localid="1654005825158" $H_{2} {SO}_{4} (a q) + Al (OH)_{3} (s) 鉄� H_{2} O (l) + {Al}_{2} {({SO}_{4})}_{3} (a q)$

Identify the weaker acid in each of the following pairs:

a. ${HNO}_{3}$ or ${HCO}_{3}^{-}$

b. ${HSO}_{4}^{-}$ or $H_{2} O$

c. $H_{2} {SO}_{4}$ or $H_{2} {CO}_{3}$

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