Question 5 of 32 Donald McQuarrie-Peter Rock . Ethan Gallogly : presented by Sap

Question

Question 5 of 32 Donald McQuarrie-Peter Rock . Ethan Gallogly : presented by Sapling Learning Mapa Phosphoric acid is a triprotic acid (K1 6.9x 103,K2 6.2x 10, and Ka3 4.8x 10-3). To find the pH of a buffer composed of H2PO4 (aq) and HPO4 (aq), which pKa value would you use in the Henderson-Hasselbalch equation? O pKa 2.16 O pKre 7.21 O pKa= 12.32 Calculate the pH of a buffer solution obtained by dissolving 12.0 g of KH2PO(s) and 25.0 g of Na2HPO4(s) in water and then diluting to 1.00 L Number pH= O Previous Give Up & View Solution e Check Answer 0 Next Exit Hint The Henderson-Hasselbalch equation is pH-pK,+ log tbase pHpK, + log (acid In this case, the acid is H2PO and the base is HPo

Explanation / Answer

(1)

In order to calculate the pH of buffer composed of H2PO4- and HPO4^2- we must use pKa2 = 7.21 value because

Ka2 = [HPO4^2-][H+] / [HPO4-]

Moles of H2PO4- = mass of KH2PO4 / Molar mass of KH2PO4 = 12 g / 136 g/mol = 0.088 mol

Moles of HPO4^2- = 25 g / 142 = 0.176 mol

[H2PO4-] = 0.088 mol/L

[HPO4^2-] = 0.176 mol/L

pH = pKa2 + log ( [ HPO4^2-] / [ H2PO4-] )

pH = 7.21 + log ( 0.176 / 0.088)

pH = 7.21 + 0.301 = 7.511

pH = 7.511

(2)

pOH = pKb + log [NH4Cl] / [NH3]

pKb of NH3 = 4.74

pOH = 4.74 + log 0.1/0.1

pOH = 4.74 ( log1 = 0)

pH = 14 - pOH = 14 - 4.74 = 9.26

initial moles of NH3 = molarity * volume in litres = 0.1 * 100/1000 = 0.01 mol

moles of NH4Cl = 0.1 * 100/1000 = 0.01 mol

moles of HCl = 0.1 * 9 / 1000 = 9 x 10^-4 mol

After addition of HCl the reaction which occurs is

NH3 + HCl -------> NH4Cl

moles of NH4Cl formed = 9 x 10^-4 mol ( because HCl is present in limited quantity so HCl is the limiting reagent)

new moles of NH4Cl = 0.01 + 9 x 10^-4 = 0.0109 mol

new moles of NH3 = 0.01 - 9 x 10^-4 = 0.0091 mol

total volume = 9 + 100 = 109 ml = 0.109 L

new [NH3] = 0.0091 / 0.109 = 0.0835 M

new [NH4Cl] = 0.0109 / 0.109 = 0.1 M

new pOH = 4.74 + log 0.1 / 0.0835

pOH = 4.74 + log 1.198

pOH = 4.74 + 0.0784 = 4.82

new pH = 14 - 4.82 = 9.18

change in pH = 9.26 - 9.18 = 0.08

3)

moles of NaOH = 0.1 * 9 / 1000 = 9 x 10^-4

NH4Cl + NaOH -------> NH3 + NaCl + H2O

new moles of NH4Cl = 0.01 - 9 x 10^-4 = 0.0091 mol

no.of moles of NH3 = 0.01 + 9 x 10^-4 = 0.0109 mol

new [NH4Cl] = 0.0091 / 0.109 = 0.0835 M

[NH3] =0.0109 / 0.109 = 0.1 M

pOH = 4.74 + log 0.0835 /0.1

pOH = 4.74 + log 0.835

pOH = 4.74 - 0.078 = 4.662

new pH = 14 - 4.662 = 9.338

change in pH = 9.338 - 9.26 = 0.078

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

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