Question

Question 2 (10 points) If 25.0 mL of 0.451 M NaOH solution is titrated with 0.253 MH2SO4, the flask at the endpoint will contain (besides the indicator phenolphthalein) as the principal components: sodium hydroxide, sulfuric acid, and water dissolved sodium sulfate and water sodium hydroxide, sodium sulfate, and water dissolved sodium sulfate, sulfuric acid, and water precipitated sodium sulfate and water Question 3 (10 points) A 75.0-ml sample of 0.0650 MHCN (K-62 10-10) is titrated with 0.65 M NaOH. What volume of 0.65 M NaOH is required to reach the stoichiometric point? 750 ml. 7.50 ml 3.75 ml. OBI

Answer 1

Question 2 Given - NaOH = 25.0 ml Volume Molarity of NG OH Q. 4.51 M Mear y Molarity of H₂SO = 0.25M Solution: - Write a complete balanced eaction sont Na Soa + 2H₂Ocel + 2NaOH, - H₂SO H28 is strong acid staong base. The Extration and strong bare give and NaOH is between strong acid neutral salt and Water . 04 8 Teil salt in aqueou Na, so is highly soluble salt. Mostly Badium salt are soluble alt in aqueous solution Na So A (ag) Na tag + 80 lag The flask at the endpoint will contain (ort 2) dissolved | Flank at the end point ! Sodium sulfate and water (besides to indicator phenolphthalein) as the principal components HO

Question 3:- Gwen:- volume Modacity Ka of Molarity Solution :- of HON = 75.0 ml = 0.075 L of HON = Q.0650 M HON = 6.2 X1010 of NaOH = 0.65M Worite the reaction between HCN and - HCN + NaOH & H₂O + Na CN HCN Moles of HON = Molarity & volume = 0.0650 md ly x 0.0754 a = 0.004875 mol =4.845 X 10-3 mol From the chemical equation, I mole of HON reacts with Imole of NaOH 4.875 x 103 mol of HCN reacts with 4.875 vie pind of NaOH 4.875 x 10 3 ml of NaGH is required to attain the stoichiometric point. point is Stoichiometaic the point at which reactant are completely react

Males Он - Urbutu Volume Vorene - Males of 120H required Holarity of NaOH = 4.875 x103 0.65 m/h -0.0075 L (12= 1000 ml) volume of NaOH = 7.50ml 012)7.50 ml o 3.55M 120H the stoichiometric point is required to each

Question 2: option 2, dissolved sodium sulfate and water

Question 3: option 2: 7.50mL of 0.65M NaOH is required to reach the stoichiometric point.