f6 CHEM

1)
Let a M be the molarity of HCl, and b M be the molarity of NaOH.

20.0 cm³ of NaOH neutralized 16.8cm³ of HCl.
NaOH + HCl → NaCl + H2O

No. of moles of NaOH = b x (20.0/1000) = 0.02b mol
No. of moles of HCl = a x (16.8/1000) = 0.0168a mol

0.02b = 0.0168a
b = 0.84a ...... [1]

CaCO3 reacted with excess HCl, and the unreacted HCl was neutralizedby NaOH.
CaCO3 + 2HCl → CaCl2 + H2O + CO2
NaOH + HCl → NaCl + H2O

No. of moles of CaCO3 = 1.00 / (40 + 12 + 16x3) = 0.01 mol
No. of moles of HCl reacted with CaCO3 = 0.01 x 2 = 0.02 mol
Total no. of moles of HCl = a x (40.0/1000) = 0.04a mol
No. of moles HCl reacted with NaOH = (0.04a - 0.02) mol
No. of moles of NaOH reacted with HCl = b x (25.0/1000) = 0.025b mol

0.04a - 0.02 = 0.025b ...... [2]

Put [1] into [2]:
0.04a - 0.02 = 0.025 x 0.84a
0.04a - 0.02 = 0.021a
0.019a = 0.02
a = 20/19 = 1.05

Put a = 20/19 into [1] :
b = 0.84 x (20/19) = 0.88

Molarity of HCl = 1.05 M
Molarity of NaOH = 0.88 M


2.
Let y M be the molarity of Y.

XOH in Y reacts with excess H2SO4, and NaOH neutralizes theunreacted acid.
2XOH + H2SO4 → X2SO­4+ 2H2O
2NaOH + H2SO4 → Na2SO­4+ 2H2O

No. of moles of NaOH = 0.4 x (12.5/1000) = 0.005 mol
No. of moles of H2SO4 reacted with NaOH = 0.005 x (1/2) =0.0025 mol
Total no. of moles of H2SO4 added = 0.1 x (100/1000) =0.01 mol
No. of moles of H2SO4 reacted with XOH = 0.01 - 0.0025 =0.0075 mol
No. of moles of XOH = 0.0075 x 2 = 0.015 mol
Molarity of XOH in Y = 0.015 / (10/1000) = 1.5 M

To make 1 dm³ of 0.1 M XOH :
No. of moles of XOH needed = 0.1 x 1 = 0.1 mol
Volume of Y needed = 0.1 / 1.5 = 0.0667 dm³ = 66.7 cm³