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Question: How many grams of sodium acetate, \(Na{{C}_{2}}{{H}_{3}}{{O}_{2}}\) , would have to be added to 1.00...

How many grams of sodium acetate, NaC2H3O2Na{{C}_{2}}{{H}_{3}}{{O}_{2}} , would have to be added to 1.00 L of 0.15 M acetic acid, HC2H3O2H{{C}_{2}}{{H}_{3}}{{O}_{2}} , to make a buffer with a pH of 5.00?

Explanation

Solution

Before solving this question, first we have to understand what is pH. pH stands for power of hydrogen and it is written as small “p” followed by a capital “H”.

Complete step by step answer:
- pH is the scale which is used to specify the acidity and basicity of a solution. The acidic solutions generally have lower pH value while the basic solutions have higher pH value of the pH scale.
To solve this question, we will use the buffer equation:
pH=pKa+log[A][HA]pH=p{{K}_{a}}+\log \dfrac{[{{A}^{-}}]}{[HA]}
Where, the value of pKap{{K}_{a}} for acetic acid is = 4.76 and the value of pH given in the question is = 5.
The concentration of acetic acid i.e. [HOAc] = 0.15M
hence, by putting the value in the above equation, we get:

& 5=4.76+\log \dfrac{[{{A}^{-}}]}{0.15} \\\ & \log \dfrac{[{{A}^{-}}]}{0.15}=+0.24 \\\ & \dfrac{[{{A}^{-}}]}{0.15} = {{10}^{0.24}} = 1.74 \\\ & [{{A}^{-}}] = [Ac{{O}^{-}}]= 0.15\times 1.74mol{{L}^{-1}} \\\ & =0.261mol{{L}^{-1}} \\\ \end{aligned}$$ Mass of sodium acetate will be$ = 1L\times 0.261mol{{L}^{-1}}\times 82.03gmo{{l}^{-1}} = 21.39g$ Hence, Mass of sodium acetate = 21.39g **Note:** The pH scale generally ranges from 0 to 14. It helps us to determine how acidic or basic the solution is. If the pH of the solution is less than 7, then it is acidic and if the pH of the solution is more than 7, then it is the basic solution. Similarly, if the pH of the solution of equal to 7, then the solution is neutral. The formula for calculation if pH is = $$pH = -\log [{{H}^{+}}]$$ \- Similarly, for a base we can calculate the concentration of OH ions using molarity and the calculate pH using the value of pOH.