50 ml of 1n ch3cooh mixed with 10 of 1 n ch3coona solution wil have an approximate ph (pka of ch3cooh=4.7 and log2=0.3)
a) 4
b) 5
c) 3
d) 7

The solution of a weak acid and its salt with strong base will make a buffer solution

Buffer solution is the solution which resists change in pH on addition of small amount of acid or base

the pH of buffer solution can be calculated using Hendersen Hassalbalch's equation

the equation is

given

pKa = 4.7

[salt] = moles of salt / total volume

moles of salt = molarity X volume = 1 X 10 = 10mmol

[salt]= 10/60 = 0.167

[acid] = moles of acid / total volume

moles of acid = molarity X volume = 1 X 50 = 50mmol

[acid]=50/60 = 0.833

pH = 4.7 + log [0.833/00167] = 4.7-0.698 =4

pH will be 4

thats how its supposed to be