A supply of NaOH contains the contaminants NaCl and MgCl2. A 4.9995g sample of this material is dissolved and diluted to 500mL, 20mL of this is titrated with 22.26mL of a 0.1989M solution of HCl. What is the percent NaOH of the original solution?

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A supply of NaOH contains the contaminants NaCl and MgCl2. A 4.9995g sample of this material is dissolved and diluted to 500mL, 20mL of this is titrated with 22.26mL of a 0.1989M solution of HCl. What is the percent NaOH of the original solution?

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Answer 1 · 2015-05-06T06:43:47

88.5%

22.26 ml of 0.1989M HCl contain $0.02226 \cdot 0.1989 = 4.427 \cdot 10^{- 3}$ $0.02226*0.1989= 4.427*10^-3$ moles of HCl. Each mole of HCl titrate 1 mole of NaOH. If 20 ml of the solution contains $4.427 \cdot 10^{- 3}$ $4.427*10^-3$ moles, in 500 ml there were 0.1107 moles of NaOH. The weight of NaOH in the sample is $0.1107 \cdot 40 (M M o f N a O H) = 4.4275$ $0.1107*40(MM of NaOH)=4.4275$ g
4.9995 g of crude material contain 4.4275 g of NaOH, therefore $\frac{4.4275}{4.9995} \cdot 100 = 88.55 %$ $4.4275/4.9995*100= 88.55%$

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A supply of NaOH contains the contaminants NaCl and MgCl2. A 4.9995g sample of this material is dissolved and diluted to 500mL, 20mL of this is titrated with 22.26mL of a 0.1989M solution of HCl. What is the percent NaOH of the original solution?

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