Titration qu causing trouble... (1 Viewer)

[VenomP]

5.267 anhydrous sodium carbonate was dissolved in water in a volumetric flask and the volume made up to 250 mL. 10 mL of this solution was pipetted into a concial flask, and titrated with hydrochloric acid. 21.3 mL was needed to reach the equivalence point.

Calculate the molarity of the hydrochloric acid solution.

 

[Pwnage101]

this is a pretty straight forward question, to be honest.

ill briefly explain how to do it

now im assuming u mean '5.267g of anhydrous sodium carbonate...'

if so, then use n=cv rearrange, and get c=n/v (work out n using n=m/M, and we are given v)

now new v is 10mL = 0.01L, While c stays the same

so now using n=cv, we work out how many moles of Na2CO3 is used up

since ratio of moles is 2:1, we need half this amount of HCL moles, and we are given v=0.0213L, so using c=n/v again using htese values should yield the result

 

[brenton1987]

5.267 anhydrous sodium carbonate was dissolved in water in a volumetric flask and the volume made up to 250 mL. 10 mL of this solution was pipetted into a concial flask, and titrated with hydrochloric acid. 21.3 mL was needed to reach the equivalence point.

Calculate the molarity of the hydrochloric acid solution.

Blue is known data. Black is unknown.

m_Na2CO3 = 5.267 g
F_Na2CO3 = 105.9887 g mol^-1
n_Na2CO3 = 0.0497 mol
V_Na2CO3 = 0.250 mL
C_Na2CO3 = 0.1988 mol L^-1

C_Na2CO3 = 0.1988 mol L^-1
V_Na2CO3 = 0.010 mL
n_Na2CO3 = 1.988*10^-3 mol

Na₂CO₃ + 2 HCl --> 2 NaCl + CO₂ + H₂O

n_Na2CO3 = 1.988*10^-3 mol
n_HCl = 3.976*10^-3 mol
V_HCl = 0.0213 L
C_HCl = 0.1866 mol L^-1

 

[x.Exhaust.x]

Blue is known data. Black is unknown.

m_Na2CO3 = 5.267 g
F_Na2CO3 = 105.9887 g mol^-1
n_Na2CO3 = 0.0497 mol
V_Na2CO3 = 0.250 mL
C_Na2CO3 = 0.1988 mol L^-1

C_Na2CO3 = 0.1988 mol L^-1
V_Na2CO3 = 0.010 mL
n_Na2CO3 = 1.988*10^-3 mol

Na₂CO₃ + 2 HCl --> 2 NaCl + CO₂ + H₂O

n_Na2CO3 = 1.988*10^-3 mol
n_HCl = 3.976*10^-3 mol
V_HCl = 0.0213 L
C_HCl = 0.1866 mol L^-1

Very nice working!