Ask Question
3 November, 11:34

Calculate the ph of a solution prepared by mixing 40.0 ml of a 0.02 m hcl solution with 200.0 ml of 0.20 m hcn solution. assume volumes to be additive ka for hcn=1.0x10^-10

+2
Answers (1)
  1. 3 November, 13:58
    0
    First, we have to get moles of each HCl & HCN

    moles of HCl = volume * molarity

    = 0.04 L * 0.02 M = 0.0008 moles

    moles of HCN = volume * molarity

    = 0.2 L * 0.2 M = 0.04moles

    ∴Total moles = 0.04 + 0.0008 = 0.0408 moles

    and the total volume = 0.04 L + 0.2 L = 0.24 L

    ∴the total concentration = total moles / total volume

    = 0.0408 moles / 0.24 L

    = 0.17 M

    when we assume [H+] = X

    ∴ Ka = X^2 / (0.17-X)

    by substitution

    ∴ 1 x 10^-10 = X^2 / (0.17 - X)

    ∴X = 4.12 x 10^-6

    ∴[H+] = 4.12 x 10^-6

    ∴PH = - ㏒ (4.12 x 10^-6)

    = 5.4
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Calculate the ph of a solution prepared by mixing 40.0 ml of a 0.02 m hcl solution with 200.0 ml of 0.20 m hcn solution. assume volumes to ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers