Question

19) A 36,3 mL aliquot of 0.0529 MH,SO, (aq) is to be titrated with 0.0411 M NaOH(aq). What volume (ml) of base will it take to reach the equivalence point? A) 93.4 B) 46.7 C) 187 D) 1.92 E) 3.84 20) What volume (mL) of a concentrated solution of sodium hydroxide (6.00 M) must be diluted to 200.0 mL to make a 0.880 M solution of sodium hydroxide? A) 2.64 B) 176 C) 26.4 D) 29.3 E) 50,0

please answer all questions and show work?

Answer 1

2 H₂SO4 + 2NaOH - Na₂SO4 + 2H₂O 1 mol of H₂SOA reacts with 2 mol of NaOH 36.3 mL of 0.0529 M of H₂SO4 = (0.0529x36,3x10 mot = 1.92810-3 mol of H₂SO4 .: 1.92 x 103 mol of H₂SO4 reacts with (2x 1.92x103 D = 3.84x 10-3 mol of Nach volume of required NaOH = numbers of moles molamity 3.84 x 10-3 mol 0.04 11 mol/L = 0.0934 L = 0.0934 x 103 mL =193.4 mL 00) Initial Molarsity (M, ) = 6.00 M Initial volume (v.) = ? final volume (V2) = 200 mL Final molarsity (M2) = 0.880 M M, V = M2N2 » V = M2 V2 0.880 M X 200 mL 6.00 M v = 29.3 mL .: Required volume of NaOH = 29.3 mL MS M,