0.25 mole of hydrogen chloride was dissolved in distilled water and the volume made up of 0.50 dm3. If 15.0 cm3 of the solution requires 12.50 cm3 of aqueous sodium trioxocarbonate (IV) for neutralization, calculate the concentration of the alkaline solution
0.30 mol dm3
0.40 mol dm3
0.50 mol dm3
0.60 mol dm3
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The answer is A.
2Hcl + Na2CO3→2NaCl+H2O+Co2
2moles of HCl →1mole of Na2Co3
0.25mole→0.25/2
=0.125
0.015dm3 of HCl →0.0125dm3 of Na2Co3
0.50dm3→(0.50*0.0125)/0.015
Vol=0.41666666666'
Molar conc=no of moles/vol in dm3
=0.125/0.41666666666'
=0.30mole/dm3
C=n/V
C=0.25/0.50=0.5mol/dm³
2HCl+NaOH ------- 2NaCl+H2CO3
CaVa/CbVb=na/nb
0.5×15/Cb×12.50 =2/1
Cb=0.5×15×1/12.50×2
Cb=7.5/25
Cb=0.3mol/dm³
The answer should be D cos 0.25 moles of HCl was dissolved in 0.50dm3 to form hydrochloric acid. Hence, the molar conc of the acid will be = amount(in moles)/volume(in dm3)
=0.25/0.50
=0.50 mol/dm3
Ca Va = Cb Vb
0.50 * 15 = x * 12.50
7.5 = x * 12.50
x = 7.5/12.50
x = 0.60 mol/dm3
Therefore,the concentration of the base is 0.60 mol/dm3.
equation of reaction
2HCl + Na2CO3 ---> 2NaCl + H2CO3
Ratio of number of moles of acid to base is 2:1
nA = 2
nB = 1
CA= 0.25mol ÷ 0.50dm³ = 0.50mol/dm³
VA = 15.0cm³
VB = 12.50cm³
CB = ?
nA/nB = CAVA / CBVB
CB = nBCAVA / VBnA
CB = (1 × 0.5 × 15) ÷ (12.50 × 2)
CB = 0.3mol/dm³
Where CA = concentration of acid
CB = concentration of base
VA = volume of acid
VB = volume of base
nA = molarity of acid
nB = molarity of base
first we find the molar volume of HCL and here we were given amount in moles of HCL as 0.25mole dissolved in volume of 0.50dm^3 of water
now
molar conc "C" = Amount (n) / Volume (V) dm^3
i.e C = 0.25/0.50 = 0.5moldm^-3
hence molar conc of the acid HCL (concentration of the acid Ca) = 0.5moldm^-3
and volume of the acid used (Va)= 15.00cm^3
then volume of the base Na2CO3 added to that of the acid for neutralisation (Vb) = 12.50cm^3
and we were asked to find the molar conc if the base Cb = ?
now in other to do that we first need to find the mole ratio of the acid to base and that can be gotten from the balanced equation 
2HCL + Na2CO3→2NaCl+H2O+Co2
na=2 : nb= 1
now using the formula (CaVa)/(CbVb) = na/nb
we have that Cb = (Ca x Va x nb)/ Vb x na
= (0.5 x 15 x 1)/(12.5 x 2)
= 0.3moldm^-3
making option A correct 
using the formula M1V1=M2V2
Given data:M1=0.25mole
V1=15cm^3
M2=12.50mole
V2=?
0.25×15=12.50×V2
3.75=12.50V2
divide both sides by 12.59
3.75/12.50=12.50V2/12.50
0.3=V2
V2=0.3cm^3
1. Find the Molarity of the Hydrochloric Acid (HCl)
First, we calculate the molarity of the acid solution that was prepared.
Moles of HCl (n): 0.25\text{ mole}
Volume (V): 0.50\text{ dm}^3
Molarity (M_a): \frac{\text{moles}}{\text{volume}} = \frac{0.25}{0.50} = \mathbf{0.50\text{ mol dm}^{-3}}
2. Write the Balanced Chemical Equation
The reaction between hydrochloric acid and aqueous sodium trioxocarbonate (IV) is:
2HCl_{(aq)} + Na_2CO_{3(aq)} \rightarrow 2NaCl_{(aq)} + H_2O_{(l)} + CO_{2(g)}
Mole Ratio (n_a : n_b): 2 : 1
3. Apply the Titration Formula
We use the general titration equation to find the unknown concentration:
\frac{M_a V_a}{M_b V_b} = \frac{n_a}{n_b}
Where:
M_a (Concentration of acid) = 0.50\text{ mol dm}^{-3}
V_a (Volume of acid used) = 15.0\text{ cm}^3
V_b (Volume of alkaline solution) = 12.50\text{ cm}^3
n_a (Moles of acid from equation) = 2
n_b (Moles of base from equation) = 1
M_b (Concentration of alkaline solution) = \text{?}
Rearranging to solve for M_b:
M_b = \frac{M_a V_a n_b}{V_b n_a}
M_b = \mathbf{0.30\text{ mol dm}^{-3}}.
