0.715 g of Na2CO3 xH2O is required 20mL of semi-normal hydrochloric acid solution for complete reaction. Find the value of x

The value of x : 2

Semi-normal Hydrochloric acid solution = 0.5 N = 0.5 M

for titration :

M₁V₁n₁=M₂V₂n₂(1=HCl,2=Na₂CO₃)

MW Na₂CO₃.xH₂O=143

MW Na₂CO₃.xH₂O = MW Na₂CO₃+ MW xH₂O = 143

MW Na₂CO₃ = 106 g/mol

MW xH₂O = 18x

Equation :

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

it is the most commonly accepted one