Post a New Question

chemistry

posted by .

a buffer is prepared by combining 10.0g of NaH2PO4 wiht 150mL of 0.20M NaOH and diluting to 2.0L. What is the pH of the buffer, Ka H2PO4-= 6.2x10-8

  • chemistry -

    The reaction between the added solid and NaOH is:
    NaH2PO4 + NaOH --> Na2HPO4 + H2O
    10.0gNaH2PO4 /120.0g/mol = 0.0833 moles NaH2PO4 initially.
    In the rection, the limiting reactant is NaOH.
    (0.150 L)(0.20 mol/L) = 0.030 mol NaOH
    0.030 moles of NaOH ---> 0.030 moles Na2HPO4 produced, and 0.030 moles NaH2PO4 used up.
    After the reaction is complete, 0.0833-0.030 = 0.0533 moles of NaH2PO4 is remaining.
    In the final mixture, the acid is H2PO4-(aq). The base is HPO4^2- Use the equation:
    pH = pKa + log(Base/Acid)

  • chemistry -

    Write the equation.
    NaH2PO4 + NaOH ==> Na2HPO4 + H2O
    Convert 10 g NaH2PO4 to moles.
    Convert 150 mL of 0.2 M NaOH to moles.
    React and calculate the concn of H2PO4 and HPO4 in the 2.0 L container.
    Then pH = pKa + log[(base)/(acid)]
    Post your work if you get stuck.

  • chemistry -

    thanks yall for your help

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question