QJEEM18Ch57d9: JEE Main 2018 Chemistry - Class 11 JEEM18Ch57d9 Solution: Following four solutions are prepared by mixing different…

(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