Asked by Dorcas

24.60 of a solution A contain H2SO4 acid is titrated against 25cm^3 portion of solution B containing 1.4g of KOH per 250cm^3 calculate the: 1).concentration of B in mol/dm^3
2). concentration of A in mol/dm^3
Answered by JAC OB254
(a)
250cm^3=1.4g
1000cm^3=??
=1000/250×1.4
=5.6g/dm^3
molarity=concentration(g/dm^3)/ RFM

RFM of KOH=39+16+1=56
Molarity=5.6/56
=0.1M=0.1mol/dm^3

(b)
H2SO4(aq)+ 2KOH(aq)> K2SO4(aq)+ 2H2O(l)
moles of 25cm^3 of KOH=25/1000×0.1
=0.0025 moles
mole ratio of KOH to H2SO4 is 2:1, thus no of moles of H2SO4= 1/2×0.0025= 0.00125 moles
molarity= 0.00125/0.0246
=0.05081M =0.0508mol/dm^3
Answered by DrBob222
moles KOH = grams/molar mass = 1.4/40 = 0.035
molarity KOH = moles/dm3 = 0.035/0.250 = 0.14 M
I assume that 24.60 is 24.60 mL.
H2SO4 + 2NaOH --> Na2SO4 + 2H2O
moles NaOH used = M x dm3 = 0.14 x 0.025 = 0.0035
mols H2SO4 = mols NaOH x 1/2 = 0.0035/2 = 0.00175
M H2SO4 in moles/dm3 = 0.00175/0.0246 = ?
Answered by DrBob222
oops. My calculation is for NaOH and not KOH. My bad.
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