# 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?

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