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Chapter 10: Q. 10.42 (page 346)

Complete the following table :

Short Answer

Expert verified

i) The pH of the component is determined as basic

ii) The pH of the component is determined as neutral

iii) The pH of the component is determined as basic

Step by step solution

01

Introduction

The given is the table with components

The objective is to complete the table

02

Explanation

The pH scale is a logarithmic scale that measures the acidity of water. $[H_{3} O^{+}]$ of aqueous solutions.

Mathematically expression of $pH$ is the negative logarithmic (base 10 ) of the $[H_{3} O^{+}]$

$[H_{3} O^{+}] [O H^{-}] = 1.0 脳 10^{- 14} {K W} = [H_{3} O^{+}] [O H^{-}]$

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

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

$pH + pOH = 14$

$pH = 14 - pOH$

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

03

Explanation

(i)

$p H = 10.00 > 7.0 s o, B a s i c [H_{3} O^{+}] = 10^{- p H} [H_{3} O^{+}] = 10^{- 10} [O H^{-}] = 1.0 脳 10^{- 10} M [O H^{-}] = = 1.0 脳 10^{- 4} M$

04

Explanation

(ii)

$pH = 7.00 = 7.0$ at, Neutral

$[H_{3} O^{+}] = 10^{- p H} [H_{3} O^{+}] = 10^{- 7}$

$[{OH}^{-}] = 1.0 脳 10^{- 7} M$

$[{OH}^{-}] = = 1.0 脳 10^{- 7} M$

05

Explanation

(iii)

$[H_{3} O^{+}] = 6.4 脳 10^{- 12} M [O H^{-}] = [O H^{-}] = [O H -] = 1.5625 脳 10^{- 3} M p H = - \log [H_{3} O^{+}] p H = - \log [6.4 脳 10^{- 12}] = 11.19 p H > 7.0 s o, B a s i c$

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Most popular questions from this chapter

Using Table $10.3$ , identify the stronger acid in each of the following pairs:

a. $HBr$ or ${HNO}_{2}$

b. $H_{3} {PO}_{4}$ or ${HSO}_{4}^{-}$

c. $HCN$ or $H_{2} {CO}_{3}$

At rest, the $[H_{3} O^{+}]$ of the stomach fluid is $2.0 脳 10^{- 4} M$ . What is the $pH$ of the stomach fluid?

Why would the pH of your blood plasma decrease if you hold your breath?

Use Le Ch芒telier's principle to predict whether each of the following changes causes the system to shift in the direction of products or reactants:

$H N O_{2} (a q) + H_{2} O (l) 鈬� N O_{2}^{-} (a q) + H_{3} O^{+} (a q)$

a. adding more $H N O_{2} (a q)$

b. removing some $N O_{2}^{-} (a q)$

c. adding more $H_{3} O^{+} (a q)$

d. removing some $H N O_{2} (a q)$

Write the formula for the conjugate acid for each of the following bases:

a. ${SO}_{4}^{2 -}$

b. $C N^{-}$

c. $N H_{3}$

d. $C l O_{2}^{-}$

See all solutions

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