a. a bottle of concentrated aqueous sulfuric acid (98.079g/mol) labled 98% wt% H
ID: 76948 • Letter: A
Question
a. a bottle of concentrated aqueous sulfuric acid (98.079g/mol) labled 98% wt% H2SO4 has aconcentration of 18.0 M NaOH. a)How many milliliters of reagentshould be diluted to 1.00 L to give 1.00 MH2SO4? b)Calculate the density of 98.0 wt%H2SO4 b.At 25oC what are the molarH3O+ and OH- concentration in: a).0671 M sodium acetate CH3COONa? b).250 MHIO3? a. a bottle of concentrated aqueous sulfuric acid (98.079g/mol) labled 98% wt% H2SO4 has aconcentration of 18.0 M NaOH. a)How many milliliters of reagentshould be diluted to 1.00 L to give 1.00 MH2SO4? b)Calculate the density of 98.0 wt%H2SO4 b.At 25oC what are the molarH3O+ and OH- concentration in: a).0671 M sodium acetate CH3COONa? b).250 MHIO3? b.At 25oC what are the molarH3O+ and OH- concentration in: a).0671 M sodium acetate CH3COONa? b).250 MHIO3?Explanation / Answer
a)M1*V1 = M2*V2
V1 = V2*(M2/M1) = 1 L *( 1 M / 18 M) = 0.0555555556 L ~ 55.6 mL
b)
18 mol H2SO4 per liter. Also 98.0% by mass H2SO4
18 mol *98.079 g/mol = 1765.422 grams H2SO4
This must be 98% of the total mass
0.98* total mass = 1765.422 g
total mass = 1801.45102 grams total in 1 Liter ofsolution
density = 1801.45102 g/1L * (1L / 1000 cm3) = 1.80 g/cm3
____________________________________________________________________ CH3COO- + H2O <=> CH3COOH + OH-
equil 0.0671 -x +x +x
Kb = [CH3COOH][OH-]/[CH3COO-] ~ x*x/(0.0671 - x) ~ x^2 / 0.0671 =1e-14/1.8e-5 =
x = (0.0671*1e-14/1.8e-5)^0.5 = 6.11e-6 M = [OH-]
[H+] = 1e-14/[OH-] = 1.64e-9 M
b)
HIO3 <=> H+ + IO3-
equil 0.250 -x x x
Ka = 10^-0.75 = x*x/(0.25 - x) = 0.178
x^2 + 0.178 x - 0.0445 = 0
x = 0.139956328 M
[H+] = x = 0.14 M
[OH-] = Kw/[H+] = 1e-14/[H+] = 7.15 e-14 M
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