JEE MAIN - Chemistry (2018 - 15th April Evening Slot - No. 19)
Following four solutions are prepared by mixing different volumes of NaOH and HCl of different concentrations, pH of which one of them will be equal to 1 ?
100 mL $${M \over {10}}$$ HCl + 100 mL $${M \over {10}}$$ NaOH
75 mL $${M \over {5}}$$ HCl + 25 mL $${M \over {5}}$$ NaOH
60 mL $${M \over {10}}$$ HCl + 40 mL $${M \over {10}}$$ NaOH
55 mL $${M \over {10}}$$ HCl + 45 mL $${M \over {10}}$$ NaOH
Explanation
(a) 100 mL $${M \over {10}}$$ NaOH will nutalise
100 mL $${M \over {10}}$$ HCl, so number extra
HCl will remain.
This will be neutral solution
$$\therefore\,\,\,$$ pH = 7
(b) Here 25 mL $${M \over 5}$$ NaOH nutralise
25 mL $${M \over 5}$$ HCl
$$\therefore\,\,\,$$ Extra HCl = 75 $$-$$ 25 = 50 mL
Total volume = 75 + 25 = 100 mL
$$\therefore\,\,\,$$ Milimole of HCl = $${{50} \over 5}$$ = 10
$$\therefore\,\,\,$$ Concentration of HCl = $${{10} \over {100}}$$ = 0.1
$$\therefore\,\,\,$$ pH = $$-$$ log[H+] = $$-$$ log(0.1) = 1
(c) HCl left = 60 $$-$$ 40 = 20 mL
$$\therefore\,\,\,$$ milimole of HCl = $${{20} \over {10}}$$ = 2
$$\therefore\,\,\,$$ Concentration of HCl = $${2 \over {100}}$$ = 0.02 M
$$\therefore\,\,\,$$ pH = $$-$$ log (0.02) = 1.69
(d) HCl left = 55 $$-$$ 45 = 10 mL
$$\therefore\,\,\,$$ milimole of HCl = $${{10} \over {10}}$$ = 1
$$\therefore\,\,\,$$ Concentration of HCl = $${{1} \over {100}}$$ = 0.01 M
$$\therefore\,\,\,$$ pH = $$-$$ log (0.01) = 2
100 mL $${M \over {10}}$$ HCl, so number extra
HCl will remain.
This will be neutral solution
$$\therefore\,\,\,$$ pH = 7
(b) Here 25 mL $${M \over 5}$$ NaOH nutralise
25 mL $${M \over 5}$$ HCl
$$\therefore\,\,\,$$ Extra HCl = 75 $$-$$ 25 = 50 mL
Total volume = 75 + 25 = 100 mL
$$\therefore\,\,\,$$ Milimole of HCl = $${{50} \over 5}$$ = 10
$$\therefore\,\,\,$$ Concentration of HCl = $${{10} \over {100}}$$ = 0.1
$$\therefore\,\,\,$$ pH = $$-$$ log[H+] = $$-$$ log(0.1) = 1
(c) HCl left = 60 $$-$$ 40 = 20 mL
$$\therefore\,\,\,$$ milimole of HCl = $${{20} \over {10}}$$ = 2
$$\therefore\,\,\,$$ Concentration of HCl = $${2 \over {100}}$$ = 0.02 M
$$\therefore\,\,\,$$ pH = $$-$$ log (0.02) = 1.69
(d) HCl left = 55 $$-$$ 45 = 10 mL
$$\therefore\,\,\,$$ milimole of HCl = $${{10} \over {10}}$$ = 1
$$\therefore\,\,\,$$ Concentration of HCl = $${{1} \over {100}}$$ = 0.01 M
$$\therefore\,\,\,$$ pH = $$-$$ log (0.01) = 2
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