If 36.0 mL of a 0.158 M NaOH solution is required to titrate 24.0 mL of a solution of H2SO4 what is the molarity of the H2SO4 solution H2SO4(aq) plus 2NaOH(aq) 2H2O(l) plus Na2SO4(aq)?

Answer 1

I think you just use C1V1 = C2V2 , but don't forget the stoichiometric ratio of

1 NaOH <=> 0.5 H2SO4

NaOH:

C1 = .1650 mol/L

V1 = 26.48 mL

H2SO4:

C2 = ? (mol/L of H2SO4)

V2 = 28.22 mL

26.49 x .1650 (mmoles of OH-) = 2 * C2 x 28.22 (mmoles of H+ from H2SO4) =

C2 = 0.5 * 0.1548848335 = 0.07744 (mol/L H2SO4)

Answer 2

Balanced equation is2NaOH + H2SO4 ==> Na2SO4 + 2H2O

mmoles H2SO4 = 86 ml x 1.75 mmoles/ml = 150.5 mmoles

mmoles NaOH needed = 150.5 mmol H2SO4 x 2 mmol NaOH/mmol H2SO4 = 301 mmole NaOH

Volume NaOH needed = 2 mmol/ml (x ml) = 301 mmole and x = 150.5 ml = 150 ml (2 sig. figs.)

Answer 3

0.062M. CALCULATION"CAVA|CBVB=NA|NB "CA=CBVBNA|VANB=0.165*26.73*1|35.46*2 =4.41045|70.92 =0.062M

Answer 4

15 ml