Question

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

Answer 1

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)

Answer 2

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.

Answer 3

thanks yall for your help