\(\mathrm{HNO}_{3}\) is a strong acid and dissolve in water gives
\(\mathrm{HNO}_{3(\mathrm{aq})} \rightarrow \mathrm{H}^{+}+\mathrm{NO}_{3}{ }^{-}\)
Each Nitric acid gives one H⁺ ions in water. So 1.0 x 10^-4 molar solution of HNO₃ gives
1.0 x 10^-4 moles of ions in water.
Therefore \(\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-4}\)
\(\mathrm{pH} =-\log _{10}\left[\mathrm{H}^{+}\right]  \)
\(=-\log _{10} 1.0 \times 10^{-4}  \)
\(=-(-4) \log _{10} 1.0 \)
\(=4 \log _{10} 10=4 \times 1 \)
pH = 4
pH of 1.0 x 10^-4 molar solution of HNO₃ is 4

NaOH is a strong base and dissociates in its solution as:
NaOH_(aq) → Na⁺_(aq) + OH^–_(aq)
One mole of NaOH would give one mole of OH^– ions. Therefore,
[OH^–] = 1 × 10^–4 mol litre^–1
pOH = –log₁₀[OH^–] = –log₁₀ × [10^–4]
= –(–4 × log₁₀10)= –(–4) = 4
Since, pH + pOH = 14
pH = 14 – pOH = 14 – 4
= 10

HCl is a strong acid and is completely dissociated in its solutions according to the process:
HCl_(aq) → H+_(aq) + Cl^–_(aq)
From this process it is clear that one mole of HCl would give one mole of H⁺ ions. Therefore, the concentration of H⁺ ions would be equal to that of HCl, i.e., 0.001 molar or 1.0 × 10^–3 mol litre^–1
Thus, [H⁺] = 1 × 10^-3 mol litre^–1.
pH = –log₁₀[H⁺] = –log₁₀ 10^–3
= –(–3 × log₁₀) = –(3 × 1) = 3
Thus, pH = 3

pH + pOH = 14
pOH = 14 – 4.5 = 9.5
pOH = 9.5

(i) No, because Nickel is more reactive than Copper.
(ii) So Nickel easily reacts and displaces copper from copper sulphate solution.
(iii) \(\mathrm{Ni}_{(\mathrm{s})}+\mathrm{CuSO_4}_{(\mathrm{aq})} \rightarrow \mathrm{NiSO_4}_{(\mathrm{aq})}+\mathrm{Cu}_{(\mathrm{S})}\)