KOH is a strong base and dissolve in water and gives
\(\mathrm{KOH_{(aq)}} \rightarrow \mathrm{K}^{+}_{(s)}+\mathrm{OH}^{-}_{(aq)}\)
Each KOH molecules gives one OH^- ion. So 1.0 x 10^-5 molar molar solution of KOH gives 1.0 x 10^-5  OH^- ions.
\({\left[\mathrm{OH}^{-}\right] } =1.0 \times 10^{-5}  \)
\(\mathrm{pOH} =-\log _{10}\left[\mathrm{OH}^{-}\right]  \)
\(=-\log _{10} 1.0 \times 10^{-5}  \)
\(=-(-5) \log _{10} 10  \)
\(\mathrm{pOH} =5 \times 1  \)
\(\mathrm{pOH} =5  \)
\(\mathrm{pH}+\mathrm{pOH} =14  \)
\(\mathrm{pH} =14-\mathrm{pOH} \)
=14 - 5 
pH = 9
The pH of 1.0 x10^-5 molar solution of KOH is 9.

Concentration of hydrogen ion H⁺ = 10^-pH M
Concentration of hydrogen ion in lemon juice = 10^-2 M
= 0.01 M
Concentration of lemon juice is 0.01 M

Here, although the solution is extremely dilute, the concentration given is not of an acid or a base but that of H⁺ ions. Hence, the pH can be calculated from the relation:
pH = –log₁₀[H⁺]
given [H⁺] = 1.0 × 10^–8 mol litre^–1
pH = –log₁₀10^–8 = –(–8 × log₁₀10)
= –(–8 × 1) = 8

Sulphuric acid dissociates in water as:
H₂SO_4(aq) → 2 H⁺_(aq) + SO₄^2-_(aq)
Each mole of sulphuric acid gives two mole of H⁺ ions in the solution. One litre of H₂SO₄ solution contains 5 × 10^–5 moles of H₂SO₄ which would give 2 × 5 × 10^–5 = 10 × 10^–5 or 1.0 × 10^–4 moles of H⁺ ion in one litre of the solution.
Therefore,
[H⁺] = 1.0 × 10^–4 mol litre^–1
pH = –log₁₀[H⁺] = –log₁₀10^–4 = –(–4 × log₁₀10)
= –(–4 × 1) = 4

(i) On passing ' C ' through water it becomes acidic.
(ii) Therefore the gas ' C ' must be a non-metal oxide \(\left(\mathrm{CO}_{2}\right)\).
(iii) So a solid compound must be a calcium carbonate.
(iv) It decomposes into calcium oxide and carbon dioxide. (C)
\(\mathrm{CaCO}_{3(\mathrm{~g})} \rightarrow \mathrm{CaO}_{(\mathrm{S})}+\mathrm{CO}_{2(\mathrm{~g})} \uparrow\\ \quad \mathrm{A} \quad \quad \quad \quad \mathrm{B} \quad \quad \quad \quad \mathrm{C}\)