JEE Main 2022 Chemistry question
200 mL of 0.01 M HCl is mixed with 400 mL of 0.01M . The pH of the mixture is .
Answer: (2)
Solution
400)/600 = (2 + 8)/600 = 1/60 M. So pH = −log(1/60) = 1.78.
200 mL of 0.01 M HCl is mixed with 400 mL of 0.01M . The pH of the mixture is .
400)/600 = (2 + 8)/600 = 1/60 M. So pH = −log(1/60) = 1.78.