chap-17-2011W-a

# 35 374 006 380 b calculate the ph if 0050 mol

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Unformatted text preview: [HA]) = -log(1.8×10-4) + log(0.40 / 0.35) = 3.74 + (0.06) = 3.80 (b) Calculate the pH if 0.050 mol of NaOH is added to 1.0 L of of the buffer solution above. HA(aq) H + (aq) + A - (aq) 0.350 0.350 M - 0.050M 0M 0.400M 0.050M 0.400M + 0.050M 0.3000.300- x x 0.450 + x pH = pK a + log[A - ] [HA] -4 = − log(1.8 ×10 ) + log(0.450 0.300) = 3.74 + 0.176 = 3.92 How How many moles of sodium acetate must be added to 1.50 L of 0.100 M acetic acid to form a pH = 5.00 buffer? Ka for acetic acid = 1.8×10-5 pKa = 4.74 pK pH = pKa + log ([A-]/[HA]) 5.00 = 4.74 + log (acetate/0.100) 5.00 - 4.74 = log (acetate/0.100) 0.26 = log (acetate) – (-1) 0.26-1 = -0.74 = log(acetate) Acetate = 10-0.74 = 0.18M Moles = Molarity × Liter = 0.18 M × 1.50 L = 0.27 moles acetate 0.27 Specify Specify the chemicals and proportions you would choose chemicals proportions from among these...
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## This document was uploaded on 02/26/2014.

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