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

Introduction

The given is the table with components

The objective is to complete the table

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 $[H_{3} O^{+}]$

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

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

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

$pH + pOH = 14$

$pH = 14 - pOH$

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

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

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

Explanation

(i)

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

Explanation

(ii)

$pH = 7.00 = 7.0$ at, Neutral

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

$[{OH}^{-}] = 1.0 \times 10^{- 7} M$

$[{OH}^{-}] = = 1.0 \times 10^{- 7} M$

Explanation

(iii)

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

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