1.) A quality control scientist at Dow Chemical tried to determine the purity of
ID: 797055 • Letter: 1
Question
1.) A quality control scientist at Dow Chemical tried to determine the purity of NaOH using acid base titration. He measured out 0.500 g of the NaOH sample, and dissolved it in 20 mL water. Then he used .416 mol/L HCL solution to titrate this NaOH samples and, 29.5 mL HCL was used to reach end point. Assume the impurity did not react with HCL, what is the purity of this NaOH sample?
2.)Excess sodium sulfate was added to a solution containing 1.50 g BaCL2, and 1.63 g white percipitate was collected. What is the percent yield of this reaction?
Please show work for both questions!!!
Explanation / Answer
(1) NaOH + HCl => NaCl + H2O
Moles of NaOH = moles of HCl = volume x concentration of HCl
= 29.5/1000 x 0.416 = 0.012272 mol
Mass of NaOH = moles x molar mass of NaOH
= 0.012272 x 40.00 = 0.49088 g
Percent purity = mass of NaOH/mass of sample x 100%
= 0.49088/0.500 x 100%
= 98.176% = 98.2%
(2) Na2SO4 + BaCl2 => BaSO4 + 2 NaCl
Moles of BaSO4 = moles of BaCl2 = mass/molar mass of BaCl2
= 1.50/208.23 = 0.0072036 mol
Theoretical yield = moles x molar mass of BaSO4
= 0.0072036 x 233.39 = 1.6812 g
Percent yield = actual yield/theoretical yield x 100%
= 1.63/1.6812 x 100%
= 96.95% = 97.0%
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