JEE Main Chemistry question
50 mL of 2 N acetic acid mixed with 10 mL of 1N sodium acetate solution will have an approximate pH of (Ka=10⁻⁵):
Answer: (A)
Solution
meq acid=100, meq salt=10. pH=pKa+log(10/100)=5−1=4.
50 mL of 2 N acetic acid mixed with 10 mL of 1N sodium acetate solution will have an approximate pH of (Ka=10⁻⁵):
meq acid=100, meq salt=10. pH=pKa+log(10/100)=5−1=4.