Question

could i see the steps for these 3 questions ?

thank you !

Acid/Base Titration Problems 6. 20.0 mL of HNO, is titrated with 0.10 M NaOH. After adding 70.0 mL of NaOH, it turned pink. What is the molarity of nitric acid? 7. Ascorbic acid is titrated to an endpoint by 14.5 mL of 0.467 M NaOH. How many mg of ascorbic acid are in each pill? (MW of ascorbic acid = 176.1 g/mol) *Hint: ascorbic acid is monoprotic 8. 10.0 mL of H3PO, is titrated with 0.501 M NaOH. 14.2 mL of NaOH was added to reach the endpoint. What is the molarity (M) of H3PO,?

Answer 1

Volume of HND, = 20.0ml = 0.020 L Volume of NaOH = 70.0mL = 0.070L Conienbation of NaOH = 0.10M = D.Imol/L no of mode of Nadu 0.070LX 0.1 mol = 7.0*16 mol Hence, no of mole of HNO, required to neutralize Naph = 7.0x15 nol. Molarity of HNO, no. of mol of HNO - 7.0x103 mob. Volume of HNO . 0.02 L = 0.35 mol/L = 0.35m. Volume of NaoH = 14.5 mL = 0.0145 L concentration of Nadh = 0.467 M = 0.467 mol/L no of male of NaOH = 0.014st 0.467 mol/ = 6.7115 xw mal no of male of ascorbic and required to neutralize Naoy = 6.7715 x 16 mol maus op Ascorbic and = nx Malon may = 6.7715x16mal x 1767 = 1.2g = 1.2x103 mg Hence, 1.2x10 mg of ascorbic and are in each pill. g/mol 87 volume of H₂PO4 = 10.omL = 0.01L Coner of NaOH = 0.5oim = o.solmol/L Volume of Nach = 14.2mL = 0.0142 no. of mode of NaOH = 0.5ol mol X 0.01421 = 7.1142 X 10ml no of male of H₂PO4 = 7.1142 x 10 mal Las H₂PO, have 3 it with Ytact with 3 NaOH) Molarity of H₂PO4 = 7:11M2 x 103 m 1 mol = 0.23714 M = 0.24 M 3x 0.01L