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Question: Following solutions were prepared by mixing different volumes of and of different concentrations. ...

Following solutions were prepared by mixing different volumes of and of different concentrations.

  1. 60 mL M10 HCl+40 mL M10 NaOH60{\text{ mL }}\dfrac{M}{{10}}{\text{ HCl}} + 40{\text{ mL }}\dfrac{M}{{10}}{\text{ NaOH}}
  2. 55 mL M10 HCl+45 mL M10 NaOH55{\text{ mL }}\dfrac{M}{{10}}{\text{ HCl}} + 45{\text{ mL }}\dfrac{M}{{10}}{\text{ NaOH}}
  3. 75 mL M5 HCl+25 mL M5 NaOH75{\text{ mL }}\dfrac{M}{5}{\text{ HCl}} + 25{\text{ mL }}\dfrac{M}{5}{\text{ NaOH}}
  4. 100 mL M10 HCl+100 mL M10 NaOH{\text{100 mL }}\dfrac{M}{{10}}{\text{ HCl}} + 100{\text{ mL }}\dfrac{M}{{10}}{\text{ NaOH}}
    pH of which one of them will be equal to 1?
    A) (ii)
    B) (i)
    C) (iv)
    D) (iii)
Explanation

Solution

To solve this first we need to calculate the number of millimoles of NaOH{\text{NaOH}} and HCl{\text{HCl}} and neutralise them. Then we can calculate the concentration using the amount of acid and base formed in the total volume. Thus, we can calculate the pH using the concentration.

Formula Used:
pH=−log⁡[H+]{\text{pH}} = - \log [{{\text{H}}^ + }]

Complete step by step answer:
Consider 60 mL M10 HCl+40 mL M10 NaOH60{\text{ mL }}\dfrac{M}{{10}}{\text{ HCl}} + 40{\text{ mL }}\dfrac{M}{{10}}{\text{ NaOH}},
Millimoles of HCl{\text{HCl}} =60 mL×M10=6 millimoles = 60{\text{ mL}} \times \dfrac{M}{{10}} = 6{\text{ millimoles}}
And,
Millimoles of NaOH{\text{NaOH}} =40 mL×M10=4 millimoles = 40{\text{ mL}} \times \dfrac{M}{{10}} = 4{\text{ millimoles}}
Thus, we have6 millimoles6{\text{ millimoles}} of HCl{\text{HCl}} and 4 millimoles4{\text{ millimoles}} of NaOH{\text{NaOH}}. 4 millimoles4{\text{ millimoles}} of HCl{\text{HCl}} will neutralise 4 millimoles4{\text{ millimoles}} of NaOH{\text{NaOH}}. Thus, the resulting solution formed will have 2 millimoles2{\text{ millimoles}} of HCl{\text{HCl}}.
The total volume of the solution is =60 mL NaOH+40 mL HCl=100 mL = 60{\text{ mL NaOH}} + 40{\text{ mL HCl}} = 100{\text{ mL}}.
The concentration of H+{{\text{H}}^ + } ions in the solution is,
[H+]=2 millimoles100 mL=0.02 M[{{\text{H}}^ + }] = \dfrac{{2{\text{ millimoles}}}}{{100{\text{ mL}}}} = 0.02{\text{ M}}
Thus,
pH=−log⁡[0.02 M]{\text{pH}} = - \log [0.02{\text{ M}}]
pH=1.69{\text{pH}} = 1.69

Thus, the pH of the solution containing 60 mL M10 HCl+40 mL M10 NaOH60{\text{ mL }}\dfrac{M}{{10}}{\text{ HCl}} + 40{\text{ mL }}\dfrac{M}{{10}}{\text{ NaOH}} is 1.69{\text{1}}{\text{.69}}.
Consider 55 mL M10 HCl+45 mL M10 NaOH55{\text{ mL }}\dfrac{M}{{10}}{\text{ HCl}} + 45{\text{ mL }}\dfrac{M}{{10}}{\text{ NaOH}},
Millimoles of HCl{\text{HCl}} =55 mL×M10=5.5 millimoles = 55{\text{ mL}} \times \dfrac{M}{{10}} = 5.5{\text{ millimoles}}
And,
Millimoles of NaOH{\text{NaOH}} =45 mL×M10=4.5 millimoles = 45{\text{ mL}} \times \dfrac{M}{{10}} = 4.5{\text{ millimoles}}
Thus, we have5.5 millimoles5.5{\text{ millimoles}} of HCl{\text{HCl}} and 4.5 millimoles4.5{\text{ millimoles}} of NaOH{\text{NaOH}}. 4.5 millimoles4.5{\text{ millimoles}} of HCl{\text{HCl}} will neutralise 4.5 millimoles4.5{\text{ millimoles}} of NaOH{\text{NaOH}}. Thus, the resulting solution formed will have 1 millimoles{\text{1 millimoles}} of HCl{\text{HCl}}.

The total volume of the solution is =55 mL NaOH+45 mL HCl=100 mL = 55{\text{ mL NaOH}} + 45{\text{ mL HCl}} = 100{\text{ mL}}.
The concentration of H+{{\text{H}}^ + } ions in the solution is,
[H+]=1 millimoles100 mL=0.01 M[{{\text{H}}^ + }] = \dfrac{{1{\text{ millimoles}}}}{{100{\text{ mL}}}} = 0.01{\text{ M}}
Thus,
pH=−log⁡[0.01 M]{\text{pH}} = - \log [0.01{\text{ M}}]
pH=2{\text{pH}} = 2

Thus, the pH of the solution containing 55 mL M10 HCl+45 mL M10 NaOH55{\text{ mL }}\dfrac{M}{{10}}{\text{ HCl}} + 45{\text{ mL }}\dfrac{M}{{10}}{\text{ NaOH}} is 2{\text{2}}.
Consider 75 mL M5 HCl+25 mL M5 NaOH75{\text{ mL }}\dfrac{M}{5}{\text{ HCl}} + 25{\text{ mL }}\dfrac{M}{5}{\text{ NaOH}},
Millimoles of HCl{\text{HCl}} =75 mL×M5=15 millimoles = 75{\text{ mL}} \times \dfrac{M}{5} = 15{\text{ millimoles}}
And,
Millimoles of NaOH{\text{NaOH}} =25 mL×M5=5 millimoles = 25{\text{ mL}} \times \dfrac{M}{5} = 5{\text{ millimoles}}
Thus, we have15 millimoles15{\text{ millimoles}} of HCl{\text{HCl}} and 5 millimoles5{\text{ millimoles}} of NaOH{\text{NaOH}}. 5 millimoles5{\text{ millimoles}} of HCl{\text{HCl}} will neutralise 5 millimoles5{\text{ millimoles}} of NaOH{\text{NaOH}}. Thus, the resulting solution formed will have 10 millimoles{\text{10 millimoles}} of HCl{\text{HCl}}.
The total volume of the solution is =75 mL NaOH+25 mL HCl=100 mL = 75{\text{ mL NaOH}} + 25{\text{ mL HCl}} = 100{\text{ mL}}.
The concentration of H+{{\text{H}}^ + } ions in the solution is,
[H+]=10 millimoles100 mL=0.1 M[{{\text{H}}^ + }] = \dfrac{{10{\text{ millimoles}}}}{{100{\text{ mL}}}} = 0.1{\text{ M}}
Thus,
pH=−log⁡[0.1 M]{\text{pH}} = - \log [0.1{\text{ M}}]
pH=1{\text{pH}} = 1

Thus, the pH of the solution containing 75 mL M5 HCl+25 mL M5 NaOH75{\text{ mL }}\dfrac{M}{5}{\text{ HCl}} + 25{\text{ mL }}\dfrac{M}{5}{\text{ NaOH}} is 1{\text{1}}.
Thus, the solution prepared by mixing 75 mL M5 HCl+25 mL M5 NaOH75{\text{ mL }}\dfrac{M}{5}{\text{ HCl}} + 25{\text{ mL }}\dfrac{M}{5}{\text{ NaOH}} will have pH equal to 1.

Thus, the correct option is (d) (iii).

Note: M is the unit of concentration which is known as molarity. In the fourth option, we are given that solution is prepared by mixing 100 mL M10 HCl+100 mL M10 NaOH{\text{100 mL }}\dfrac{M}{{10}}{\text{ HCl}} + 100{\text{ mL }}\dfrac{M}{{10}}{\text{ NaOH}}. Here, equal volumes of acid and base are mixed. Also, the concentrations of acid and base are equal. Thus, the pH of the solution will be neutral.