Question

Questions 1. Calculate the molarity of a sodium hydroxide (NaOH) solution that is titrated with 0.6887 g of oxalic acid (Equation 2). The titration requires 15.80 mL of the NaOH solution to reach the end point. Calculate the molarity of a sulfuric acid (H,SO) solution if 30.10 mL of 0.62 10 M NaOH is required to reach the end point when titrated against 10.00 mL of the unknown acid solution. The balanced chemical equation for the reaction is given below. (Refer to Example 2 for help with calculations) 2. Na,SO,(aq) + 2 H,00) H,SO,(aq) +2 NaOH(aq) -

Answer 1

1.) males of oxalic acid = Mass - Do 68879 • Malau mass 90.03g/mas a males = 7. 65x10 mal C ₂ H ₂ O (aq) + 2Naoh (49) - Naz C20 g + 2100 moles of Calg On - males of Naon officient officient. - Malarity X Volume 7665x103 70658103, Molarity X 15.8X10-PL 2 Molarity = 2X7065X/03 1508X20-3 Malauity = 0.968M

2 2 Ron H₂SO4 (aq) + 2NaOH (ar) Nassou Can ) + 2H06) males of Naoh moles of He sou cofficient Molarity X 30x/03 Cofficiente - 00621 MX 30 1x102 Molauity = 0:621X30X60-3 2X10X10-3 Malarity = 0.9315M So Malouiety of Me Soy Sol"= 0.9315M Thank you s May it help you :)