Unit 10 H NH,CI after 8. What is the phl of a 1.0 L. buffer that consists of 0.0

Question

Unit 10 H NH,CI after 8. What is the phl of a 1.0 L. buffer that consists of 0.0750 M C,H,NH, and 0.0500 M C the addition of 10.0 ml. of 0.5 M HCP a. Calculate moles of C,H,NH, CH,NH,Ci, and HCl b. Write a reaction of the HCl with the appropriate other species the reaction as a header. c. Set up a reaction table (will use moles and stochiometry) using +CI (aq) HCI(aq) Before Reacts After d. Looking at your After line, use that information to determine what you have present in solution and write down what you would need to do in order to solve for the pH e. Solve for the pH! You should be able to use either an ice table or the Henderson-Hasselbach tion.

Explanation / Answer

a) Volume changes and concentration also changes, but moles never changes even on dilution

moles C2H5NH2 = 0.075 mol

moles C2H5NH3Cl = 0.05 mol

moles HCl = 10 mL x 10-3 x 0.5 M = 0.005 mol

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b)

C2H5NH2 + HCl <==> C2H5NH3+ + Cl-

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c)

......................C2H5NH2 (aq) + HCl (aq) <==> C2H5NH3+ (aq) + Cl- (aq)

Before ................. 0.075 ......... 0.005................... 0.05 ........... 0.05

Reacts................ -0.005 ........ -0.005 ................ +0.005 .........+0.005

After .................. ..0.07 .............0 ......................0.055 .............0.055

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d) Assuming the volume change after addition of 10 mL of 0.5M HCl to 1000 mL is negligible, now a weak base C2H5NH2 and its conjugate acid C2H5NH3Cl exist in solution, which makes a buffer. I need base dissociation value of ethylamine in order to calculate this buffer pH. The Kb value i found is Kb = 6.3×10–4.

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e)

From Henderson-Hasselbalch equation for basic buffer

pOH = pKb + log[conjugate acid / base]

pKb = -logKb = -log(6.3×10–4) = 3.2007

pOH = 3.2 + log[0.055 / 0.07]

pOH = 3.2007 - 0.1047 = 3.096 = 3.1

pOH = 3.1, but pH + pOH = 14

pH = 14 - 3.1 = 10.9

pH = 10.9

Hope this helped you!

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