164041
The \(\mathrm{pH}\) of the solution containing \(10 \mathrm{~mL}\) of a \(0.1 \mathrm{M}\) \(\mathrm{NaOH}\) and \(10 \mathrm{~mL}\) of \(0.05 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4\) would be:
1 Zero
2 1
3 \(>7\)
4 7
Explanation:
milli equivalent of \(\mathrm{H}^{+}\)ion \(=\mathrm{N} \times \mathrm{V}\) \( =M \times v . f \times V \) \( =0.05 \times 2 \times 10=1 \) milli equivalent of \(\mathrm{OH}^{-}=0.1 \times 10=1\) milli equivalent of \(\mathrm{H}^{+}=\)milli equivalent of \(\mathrm{OH}^{-}\) \( \therefore \mathrm{pH}=7 \)
NCERT-XI-214
5 RBTS PAPER
164042
\(8 \mathrm{~g} \mathrm{NaOH}\) and \(4.9 \mathrm{H}_2 \mathrm{SO}_4\) are present in one litre of the solution. What is its \(\mathrm{pH}\) ?
164041
The \(\mathrm{pH}\) of the solution containing \(10 \mathrm{~mL}\) of a \(0.1 \mathrm{M}\) \(\mathrm{NaOH}\) and \(10 \mathrm{~mL}\) of \(0.05 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4\) would be:
1 Zero
2 1
3 \(>7\)
4 7
Explanation:
milli equivalent of \(\mathrm{H}^{+}\)ion \(=\mathrm{N} \times \mathrm{V}\) \( =M \times v . f \times V \) \( =0.05 \times 2 \times 10=1 \) milli equivalent of \(\mathrm{OH}^{-}=0.1 \times 10=1\) milli equivalent of \(\mathrm{H}^{+}=\)milli equivalent of \(\mathrm{OH}^{-}\) \( \therefore \mathrm{pH}=7 \)
NCERT-XI-214
5 RBTS PAPER
164042
\(8 \mathrm{~g} \mathrm{NaOH}\) and \(4.9 \mathrm{H}_2 \mathrm{SO}_4\) are present in one litre of the solution. What is its \(\mathrm{pH}\) ?
164041
The \(\mathrm{pH}\) of the solution containing \(10 \mathrm{~mL}\) of a \(0.1 \mathrm{M}\) \(\mathrm{NaOH}\) and \(10 \mathrm{~mL}\) of \(0.05 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4\) would be:
1 Zero
2 1
3 \(>7\)
4 7
Explanation:
milli equivalent of \(\mathrm{H}^{+}\)ion \(=\mathrm{N} \times \mathrm{V}\) \( =M \times v . f \times V \) \( =0.05 \times 2 \times 10=1 \) milli equivalent of \(\mathrm{OH}^{-}=0.1 \times 10=1\) milli equivalent of \(\mathrm{H}^{+}=\)milli equivalent of \(\mathrm{OH}^{-}\) \( \therefore \mathrm{pH}=7 \)
NCERT-XI-214
5 RBTS PAPER
164042
\(8 \mathrm{~g} \mathrm{NaOH}\) and \(4.9 \mathrm{H}_2 \mathrm{SO}_4\) are present in one litre of the solution. What is its \(\mathrm{pH}\) ?
164041
The \(\mathrm{pH}\) of the solution containing \(10 \mathrm{~mL}\) of a \(0.1 \mathrm{M}\) \(\mathrm{NaOH}\) and \(10 \mathrm{~mL}\) of \(0.05 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4\) would be:
1 Zero
2 1
3 \(>7\)
4 7
Explanation:
milli equivalent of \(\mathrm{H}^{+}\)ion \(=\mathrm{N} \times \mathrm{V}\) \( =M \times v . f \times V \) \( =0.05 \times 2 \times 10=1 \) milli equivalent of \(\mathrm{OH}^{-}=0.1 \times 10=1\) milli equivalent of \(\mathrm{H}^{+}=\)milli equivalent of \(\mathrm{OH}^{-}\) \( \therefore \mathrm{pH}=7 \)
NCERT-XI-214
5 RBTS PAPER
164042
\(8 \mathrm{~g} \mathrm{NaOH}\) and \(4.9 \mathrm{H}_2 \mathrm{SO}_4\) are present in one litre of the solution. What is its \(\mathrm{pH}\) ?
