Question

mem%20210%20FINAL%20EXAM%20(SU20)%20.pdf Imported From Sa.. 2. A student prepared a stock solution by dissolving 15.0 g of NaOH in enough water to make 150 mL of a stock solution. She then took 22.5 ml of the stock solution and diluted it with enough water to make 250 mL of a working solution. (5 pts. each) A. What is the molar concentration of NaOH in the stock solution? B. What is the molar concentration of NaOH in the working solution? C. How much of the working solution will she need to neutralize 10.5 mL of 0.250 M H3PO4?

Answer 1

mo17. Solution : Moleculas weight of NaoH = (23+1+16) = 40 g : Number of mole of Naot = 15.0 g 40 g molt Therefore, 0.375 mol Naoh Present 150.0 ml = 0.375 mol stock solution. A. Molas Concentration of Nauh in the stock solution is 0.375 mol X 1000 2.5 (M) 1500ml Again, 150 ml stock solution contain 0.375 mol raolt. 22.5 65 и 0.375 -X 22.5 4 150 = 0.05625 mol NooH. B B. The molas concentration of mach in the world working Solution is → = 0.225 (M) 0.05625 mol 250.0m X 1000 C. neutralization formula : VpS, = V2S2 V. (volume of H₂ po y) = 10.5 mi th of H₂ poy) = 0.250 (M) V2 (volume of NaoH) = ? si ( strength Sz Cs strength of Noo) = 0.225 (M). .. Visi S2 = I1.67 mi (10.5 mi) x (0.250 m) V2= (0.225 m) Hence, 11.67 mi of the working solution will she need to neutralize 10.5 ml of 0.250 m H₂poy.