Hcl is titrated against 5ml of sodium carbonate of strength 0.0133 then what should be the mean value of HCl added ?
To find the mean value of HCl added during the titration, we need to use the concept of stoichiometry and the equation for the reaction between HCl and sodium carbonate.
The balanced equation for the reaction is:
HCl + Na2CO3 -> NaCl + H2O + CO2
From the equation, we can see that one molecule of HCl reacts with one molecule of sodium carbonate. Therefore, the ratio of HCl to sodium carbonate is 1:1.
Given that the volume of sodium carbonate solution is 5 mL and the concentration of sodium carbonate is 0.0133 M, we can calculate the number of moles of sodium carbonate:
moles of sodium carbonate = concentration * volume
= 0.0133 M * 5 mL
= 0.0665 moles
Since the ratio of HCl to sodium carbonate is 1:1, the number of moles of HCl required is also 0.0665 moles.
Now, let's assume that during the titration, it took a total volume of V mL of HCl to neutralize the sodium carbonate solution. The mean value of HCl added can be calculated as:
mean value of HCl added = total moles of HCl added / total volume of HCl added
Since the moles of HCl required is 0.0665 moles, the mean value of HCl added can be calculated as:
mean value of HCl added = 0.0665 moles / V mL
Therefore, we need to know the total volume of HCl added (V mL) to calculate the mean value of HCl added.