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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 ?

  • Option 1)

    100\;mL\;\frac{M}{10}\;HCl + 100\;mL\;\frac{M}{10}\;NaOH

  • Option 2)

    75\;mL\;\frac{M}{5}\;HCl + 25\;mL\;\frac{M}{5}\;NaOH

  • Option 3)

    60\;mL\;\frac{M}{10}\;HCl + 40\;mL\;\frac{M}{10}\;NaOH

  • Option 4)

    55\;mL\;\frac{M}{10}\;HCl + 45\;mL\;\frac{M}{10}\;NaOH

 

Answers (1)

As we have learnt,

 

The p(H) scale -

Hydronium ion concentration in molarity is more conveniently expressed on a logarithmic scale known as p(H) scale.

- wherein

The p(H) of a solution is defined as negative logarithm to base 10 of the activity of hydrogen ion

P(H)=-log[H^{+}]

 

 miliequivelent of HCl - miliequivelent of NaOH

75\times\frac{1}{5} - 25\times\frac{1}{5} =Remaining miliequivelent of HCl

15 - 5 = 10

N of remaining HCl = \frac{10}{75+25} = 10^{-1}

[H+] = 10-1

pH = -log 10-1 = 1

 


Option 1)

100\;mL\;\frac{M}{10}\;HCl + 100\;mL\;\frac{M}{10}\;NaOH

Option 2)

75\;mL\;\frac{M}{5}\;HCl + 25\;mL\;\frac{M}{5}\;NaOH

Option 3)

60\;mL\;\frac{M}{10}\;HCl + 40\;mL\;\frac{M}{10}\;NaOH

Option 4)

55\;mL\;\frac{M}{10}\;HCl + 45\;mL\;\frac{M}{10}\;NaOH

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