50   mL of 0 . 02 MNaOH solution is mixed with 50   mL of 0 . 06 M acetic acid solution, the pH of the resulting solution is_________ ( pKa of acetic acid is 4 . 76 ,   log 5 = 0 . 70 ) .

Question

50 &#160; mL of 0 . 02 MNaOH solution is mixed with 50 &#160; mL of 0 . 06 M acetic acid solution, the pH of the resulting solution is_________ ( pKa of acetic acid is 4 . 76 , &#160; log 5 = 0 . 70 ) .

Aicharya · Accepted Answer

When a strong base is added to a weak acid solution, it results in the formation of a salt. Here, acid is present in a limiting reagent and base is present in excess amounts. So, by using the pH formula:

$\begin{matrix} p H & = & p K a + \log \frac{[s a l t]}{[a c i d]} \\ p H & = & 4.46 + \log \frac{[0.02 \times 50]}{[(0.06 - 0.02) \times 50]} \\ p H & = & 4.76 + \log (0.5) \\ p H & = & 4.46 \end{matrix}$

50   mL of 0 . 02 MNaOH solution is mixed with 50   mL of 0 . 06 M acetic acid solution, the pH of…

Solution

50 &#160; mL of 0 . 02 MNaOH solution is mixed with 50 &#160; mL of 0 . 06 M acetic acid solution, the pH of…

Solution

50 mL of 0 . 02 MNaOH solution is mixed with 50 mL of 0 . 06 M acetic acid solution, the pH of…