# For titration of 5 ml of potassium carbonate solution, 13 ml of hydrochloric acid solution with C (1 / z HCl)

For titration of 5 ml of potassium carbonate solution, 13 ml of hydrochloric acid solution with C (1 / z HCl) = 0.095 mol / L was consumed. Calculate C (1 / z K2CO3), T (K2CO3) and m (K2CO3) in solution The reaction equation has the form:
K2CO3 + 2HCl = 2KCl + H2O + CO2 ↑
M (K2CO3) = 138 g / mol.
a) Calculation of C (1 / z K2CO3)
The number of moles of equivalents of a substance can be calculated through the mass m (B) of substance B:
C (1 / z K2CO3) = m (B) / M (1 / zB)
To calculate according to the titration results, we use the mathematical expression of the consequence of the law of equivalence (the so-called rule of proportionality):
CE (A). V (A) = CE (B). V (B)
Where
CE (A) and CE (B) – normal concentrations of equivalents of substances K2CO3 and HCl, mol / l; VA) and V (B) – volumes of solutions of substances K2CO3 and HCl.
Then
normality CE (K2CO3) = [CE (HCl). V (HCl)] / V (K2CO3) = (0.095. 13) / 5 = 0.247 mol / L.
b) Find the mass of K2CO3, we get:
m (K2CO3) = CE (K2CO3). [M (K2CO3) / 2]. [(5/1000)] = [0.247. (138/2) / (5/1000] = 0.085g.
c) Calculate the titer T (K2CO3), we get:
T (K2CO3) = m (K2CO3) / V (K2CO3) = 0.085 / 5 = 0.017 g / ml.
Answer: CE (K2CO3) = 0.247 mol / L; m (K2CO3) = 0.085g; T (K2CO3) = 0.017g / ml Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.