Solveeit Logo
Login

Question

Question: For \({\text{500 mL}}\) of \({\text{0}} \cdot {\text{2 M}}\) aqueous solution of acetic acid is mixe...

For 500 mL{\text{500 mL}} of 0⋅2 M{\text{0}} \cdot {\text{2 M}} aqueous solution of acetic acid is mixed with 500 mL{\text{500 mL}} of 0⋅2 M HCl{\text{0}} \cdot {\text{2 M HCl}} at 25oC{25^o}C. If 6 g{\text{6 g}} of NaOH{\text{NaOH}} is added to the above solution, determine the final pH (assuming no change in volume on mixing. Ka{K_a} of acetic acid is 1⋅75×10−5molL−11 \cdot 75 \times {10^{ - 5}}mol{L^{ - 1}}. (Write the value of nearest integer)

Explanation

Solution

First calculate the number of moles present of NaOH{\text{NaOH}}. To know the value of pH one must always first try to find the value of pKap{K_a}. Once if you know the value of pKap{K_a} you can directly substitute the values in the formulae to identify the pH.

Complete answer:

  1. First of all we will calculate the number of moles of NaOH{\text{NaOH}} as below,
    Number of moles of NaOH{\text{NaOH}} =Given weightMolecular weight of NaOH = \dfrac{{{\text{Given weight}}}}{{{\text{Molecular weight of NaOH}}}}
    Weight is given of NaOH = 6 grams{\text{NaOH = 6 grams}}
    Molecular weight of NaOH = 23 + 16 + 1 = 40 grams{\text{NaOH = 23 + 16 + 1 = 40 grams}}
    Now let us put these values in the above formula we get,
    Number of moles of NaOH{\text{NaOH}} =Given weightMolecular weight of NaOH = \dfrac{{{\text{Given weight}}}}{{{\text{Molecular weight of NaOH}}}}
    Number of moles of NaOH{\text{NaOH}} =640 = \dfrac{6}{{40}}
    Number of moles of NaOH{\text{NaOH}} =0⋅15moles = 0 \cdot 15moles
  2. Now out of 0⋅15moles0 \cdot 15moles of NaOH{\text{NaOH}} 0⋅1M{\text{0}} \cdot {\text{1M}} will be neutralized with HCl{\text{HCl}} (which is a strong acid) and 0⋅005M{\text{0}} \cdot {\text{005M}} will be neutralized with acetic acid (which is a weak acid). Hence, we can say that,
    [CH3COOH]=0⋅05M\left[ {C{H_3}COOH} \right] = 0 \cdot 05M and [CH3COONa]=0⋅05M\left[ {C{H_3}COONa} \right] = 0 \cdot 05M
  3. As mentioned before now we need to find out the value of pH and for that, we need the value of pKap{K_a} as below,
    pKa=−log⁡10[Ka]p{K_a} = - {\log _{10}}\left[ {{K_a}} \right]
    Now the value of Ka{K_a} is given as 1⋅75×10−51 \cdot 75 \times {10^{ - 5}} and lets put that value in the above equation,
    pKa=−log⁡10[1⋅75×10−5]p{K_a} = - {\log _{10}}\left[ {1 \cdot 75 \times {{10}^{ - 5}}} \right]
    After calculating the above logarithmic value we get,
    pKa=4⋅75p{K_a} = 4 \cdot 75
  4. As we know the pKap{K_a} value we can calculate the pH by using the following formula,
    pH=pKa+log⁡[CH3COOH/CH3COONa]pH = p{K_a} + \log \left[ {C{H_3}COOH/C{H_3}COONa} \right]
    By putting the known values in the above equation we get,
    pH=4⋅75+log⁡[0⋅05/0⋅05]pH = 4 \cdot 75 + \log \left[ {0 \cdot 05/0 \cdot 05} \right]
    pH=4⋅75+log⁡[1]pH = 4 \cdot 75 + \log \left[ 1 \right]
    As the value of log⁡[1]=0\log \left[ 1 \right] = 0 we get,
    pH=4⋅75+0pH = 4 \cdot 75 + 0
    pH=4⋅75pH = 4 \cdot 75
    Therefore, the pH value is 4⋅75{\text{4}} \cdot {\text{75}} and the nearest integer value is 5{\text{5}}. Hence, the pH is 5{\text{5}}.

Note:
Always remember to find out the number of moles for the given compound when gram addition is done in a solution. Finding the pKap{K_a} value before pH will make it easy to find out the value of pH.