4/8/2016 12:00 PM 7.1/364/6/2016 10:21 PM Gradebook -Print Calculator-d Periodic

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4/8/2016 12:00 PM 7.1/364/6/2016 10:21 PM Gradebook -Print Calculator-d Periodic Table O Assignmer Available Fror Due Date: Points Possibl Grade Catego Description: Policies: Question 7 of 36 University Science Books presented by Sapling Learning Mapcab Donald McQuarrie Peter A. Rock. Ethan Gallogly Phosphoric acid is a triprotic acid (Kal-6 Six 10-3, Ka2-6.2×10-8, and Ka3 = 4 8x1ors). To find the pH of a buffer composed of H2PO (aq) and HPO4- (aq), which the Henderson-Hasselbalch equation? pk, value would you use in 0 pKa1 = 2.16 pKa = 7.21 pka3-12.32 You can check You can view so up on any ques You have five at Calculate the pH of a buffer solution obtained by dissolving 23.0 g of KH2PO4(s) and 42.0 g of Na2HPO4(s) in water and then diluting to 1.00 L. You lose 5% of in your question answer Number O Help With This Web Help & Vid O Technical Supp pH = 11

Explanation / Answer

Q.1: In the buffer composed of H2PO4-(aq) and HPO42-(aq), the H2PO4-(aq) act as acid and gives H+ ion. Hence we have to use the pKa2 ( = 7.21) in the Hendersen - Hasselbalch equation.

Q.2: Given the mass of KH2PO4 = 23.0 g

molar mass of KH2PO4 = 136.08 g/mol

Hence moles of KH2PO4 = 23.0 g / 136.08 g/mol = 0.169 mol

Volume of the buffer solution, V = 1.00 L

Hence [KH2PO4] = 0.169 mol / 1.00 L = 0.169 M

Given the mass of Na2HPO4(s) = 42.0 g

molar mass of Na2HPO4(s) = 141.96 g/mol

Hence moles of Na2HPO4 = 42.0 g / 141.96 g/mol = 0.296 mol

Volume of the buffer solution, V = 1.00 L

Hence [Na2HPO4] = 0.296 mol / 1.00 L = 0.296 M

Now pH of the buffer solution can be calculated as

pH = pKa + log[Na2HPO4] / [KH2PO4]

=> pH = 7.21 + log(0.296 M / 0.169 M) = 7.45 (answer)

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