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Published on: 12/06/2021
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Questions + Answers key
Take MCQ Science Test1.
Calculate the pH of 1.0 ×10–4 molar solution of HNO3.
2.
Calculate the pH of 1 × 10–4 molar solution of NaOH.
3.
Calculate the pH of 0.001 molar solution of HCl.
4.
If the pH of a solution is 4.5, what is its pOH?
5.
Can a nickel spatula be used to stir copper sulphate solution? Justify your answer.
1.
\(\mathrm{HNO}_{3}\) 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
2.
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
3.
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
4.
pH + pOH = 14
pOH = 14 – 4.5 = 9.5
pOH = 9.5
5.
(i) No, because Nickel is more reactive than Copper.
(ii) So Nickel easily reacts and displaces copper from copper sulphate solution.
(iii) \(\mathrm{Ni}_{(\mathrm{s})}+\mathrm{CuSO_4}_{(\mathrm{aq})} \rightarrow \mathrm{NiSO_4}_{(\mathrm{aq})}+\mathrm{Cu}_{(\mathrm{S})}\)
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Tamilnadu Stateboard 10th Standard Subjects
Tamilnadu Stateboard Standards
