JEE Main 2018 — Ionic Equilibrium Question with Solution

(a)    100 mL  $$\frac{M}{10}$$ NaOH will nutalise

100 mL   $$\frac{M}{10}$$ HCl, so number extra

HCl   will remain.

This will be neutral solution

$$\therefore$$ pH = 7

(b)    Here 25 mL  $$\frac{M}{5}$$ NaOH   nutralise

25 mL  $$\frac{M}{5}$$ HCl

$$\therefore$$ Extra HCl   =   75 $$-$$ 25   =   50 mL

Total volume   =   75 + 25   =   100 mL

$$\therefore$$ Milimole of HCl   =   $$\frac{50}{5}$$ = 10

$$\therefore$$ Concentration of HCl  =  $$\frac{10}{100}$$  = 0.1

$$\therefore$$ pH  =  $$-$$ log[H⁺]  =  $$-$$ log(0.1)  =  1

(c)    HCl left  =  60 $$-$$ 40  =  20 mL

$$\therefore$$ milimole of HCl   =   $$\frac{20}{10}$$   =  2

$$\therefore$$ Concentration of HCl   =   $$\frac{2}{100}$$   =  0.02 M

$$\therefore$$ pH  =  $$-$$ log (0.02)   =   1.69

(d)    HCl left  = 55 $$-$$ 45  =  10 mL

$$\therefore$$ milimole of HCl   =  $$\frac{10}{10}$$  = 1

$$\therefore$$ Concentration of HCl  = $$\frac{1}{100}$$   =  0.01 M

$$\therefore$$ pH = $$-$$ log (0.01) = 2