1) A 0.1475M solution of Ba(OH)2 was used to titrate the acetic acid (C2H4O2=60.

Question

1) A 0.1475M solution of Ba(OH)2 was used to titrate the acetic acid (C2H4O2=60.05) in a dilute sample. The results are tabulated below. 2C2H4O2+Ba(OH)2=Ba(C2H3O2)+2H2O sample Sample Volume, mL Volume Ba(OH)2, mL 1 50.00 43.17 2 49.50 42.68 3 25.00 21.47 4 50.00 43.33 Calculate the mean and standard deviation of w/v percentage of acetic acid and the 95% confidence interval for the mean. Calcium in a mineral was analyzed five times by each of two methods. Are the mean values significantly different at the 95% confidence level? Method1: 0.0271 0.0282 0.0279 0.0271 0.0275 Method2: 0.0271 0.0268 0.0263 0.0274 0.0269

Explanation / Answer

A.HKC8H4O41.00794 + 39.0983 + (12.0107 x 8) + (1.00794 x 4) + (15.9994 x 4) =204.22 g/molB.1 mol KHP x 0.5591 g KHP = 0.002738 molKHP204.22 g KHP C.in a titration, mol KHP = mol NaOHmol KHP = 0.002738mol NaOH = mol KHP = 0.002738 mol13.39 mL = 0.01339 L0.002738mol NaOH = 0.2045 M NaOH0.01339 L NaOHD.M1V1 = M2V2(0.2045 M)(0.00838 L) = M(0.00500 L)M = 0.3427 M

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