How would you prepare 250mL of a buffer having a pH of 3.60 using 0.500M NaOH an

Question

How would you prepare 250mL of a buffer having a pH of 3.60 using 0.500M NaOH and 0.500M of HNO2? Would you calculate the molarities of the conjugate acid and base pair and then somehow get the volumes from that? Please explain

Explanation / Answer

We need to make a reaction table for a solution containing a Strong Base and a Weak Acid. It is important to realize that the reaction will be extensive with a strong base so it will completely use up the amount of NaOH in the solution. OH^1- + HNO2 ---> NO2^1- + H2O in: (8.00x) (0.1575) (0.0) (0.0) (mol) delta: (-8.00x) (-8.00x) (8.00x) (8.00x) (mol) eq: (0.0) (0.1575 - 8.00x) (8.00x) (8.00x) (mol) Note* 0.1575 = (.450 x .350) We know the equation: pH = pKa + log(nb/na) the pKa of HNO2 is 3.40 so we get: 3.30 = 3.40 + log([NO2^1-]/[HNO2]) 3.30 = 3.40 + log((8.00x)/(0.1575 - 8x)) Solve for x and notice this is the volume of NaOH needed to prepare a pH = 3.30 buffer (note: the volume will be in L since the reaction table was completed in mol and not mmol...don't forget to convert!) Answer: 8.7 ml (check me on th

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

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