[danx3]

A 0.196-g sample of a CO3^2- antacid is dissolved with 25.0 ml of0.0946M HCl .The hydrochloric acid that is not neutralized by the antacid is titrated to a bromophenol blue endpoint with 6.22ml of 0.0827M NaOH

Assuming the active ingreient in the antacid sample is CaCO3 , calculate the mass of CaCO3 in the sample (CaCO3 = 100.09g/mol)


i just hvn't finish this question , other part i hv finished
may someone help me ...thanks,

[Borek]

Pls rd fr rls.

[danx3]

Pls rd fr rls.

i have try to do

it is my answer
mole of HCl = 25/1000 x 0.0946 = 2.365x10^-3 mol

HCl + NaOH-->NaCl + H2O
mole of NaOH = 6.22/1000 x 0.0827 = 5.14394x10^-4 mol
= mole of HCl reacted with NaOH

mole of HCl reacted with CO3^2- = 2.365x10^-3 - 5.14394x10^-4
= 1.850606x10^-3 mol

CO3^2- + HCl<-->HCO3- + Cl-
HCO3- + HCl<-->H2CO3 + Cl-
----------------------------------------------
CO3^2- + 2HCl<-->H2CO3 + Cl-

<--> equilibrium

mole of CO3^2- = 1.850606x10^-3 / 2
= 9.25303x10^-4 mol
mass of CaCO3 = (9.25303x10^-4)(100.09)
= 0.0926g


may someone  help me check the answer  ? thanks

[Borek]

I have not checked numbers, but your approach seems to be correct.

[stewie griffin]

Pls rd fr rls.

Oh no! Borek, did your keyboard's vowels stop working 

[Borek]

No, I just tried to bend down to the OP level in hope my message will get through.