GATE Civil Engineering (CE) 2020 Shift 1 Solved Paper

Question : 16 of 65

Marks: +1, -0

An amount of

35.67 {mg} {HCl}

is added to distilled water and the total solution volume is made to one litre. The atomic weights of

{H}

and

{Cl}

are 1 and

35.5

, respectively. Neglecting the dissociation of water, the

{pH}

of the solution, is

Solution:

1 mole of

{HCL}

given 1 mole

{H}^{+}

ions

36.5 {gm}

{HCl}

given

1 {gm}

{H}^{+}

ions
Therefore,

35.67 {mg}

{HCl}

gives,

\;{1}/{36.5} × 36.67 {mg}

{H}^{+}

ions

∴ 35.67 {mg}

{HCl}=0.977 {mg}

{H}^{+}

\;{0.977 × 10^{-3}}/{1}

=9.77 × 10-4

moles of

{H}^{+}

ions
Now, we know that

{pH}=-\log _{10}[ {H}^{+}]

⇒ {pH}=-\log _{10}(9.77 × 10^{-4})

\;\; =-\log _{10} 9.77+(-\log _{10}(10^{-4}))

\;\; =-\log _{10} 9.77+4 \log _{10} 10

\;\; =-0.989+4

⇒ {pH}=3.01

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