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

pure water does not conduct electricity very well. however, when certain substances are dissolved in water, the solution does conduct electricity. you can make a simple device that shows how well a solution conducts electricity. this device uses a flashlight bulb to indicate how well the solution conducts electricity. the better the solution conducts electricity, the brighter the bulb will glow.

explanation:

the answer is the first option

explanation:

just googled it m8. not that hard

lol