a buffer is prepared by combining 10.0 g of NaH2PO4 with 150 mL of 0.20 M NaOH a

Question

a buffer is prepared by combining 10.0 g of NaH2PO4 with 150 mL of 0.20 M NaOH and diluting to 2.0 L. what is the pH of the buffer? (Ka (H2PO4-)= 6.2 x 10^-8)

Explanation / Answer

when NaOh reacts with the salt which is added we have the reaction NaH2PO4 + NaOH --> Na2HPO4 + H2O limiting reactant is NaOH in the reaction 10.0gNaH2PO4 /120.0g/mol = 0.0833 moles NaH2PO4 initially. (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 consumed. at the end of the reaction 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)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.