Question #67635
what is the ppm CaCO3 of a water sample if 34 mL of tap water, 26 mL of distilled water, and 3.7 mLof ETDA was used
Expert's answer
Using the law of equivalence:
c_1 V_1=c_2 V_2
The V1 is the volume of the probe: 34 mL of tap water + 26 mL of distilled water=50 mL
The c1 is the concentration of CaCO3 in M
The V2 is the volume of the EDTA in mL
The c2 is the concentration of EDTA in M
c_1=(c_2 V_2)/V_1 =(3.7 c_2)/50 M
Concentration of CaCO3 in ppm is:
c_1=(3.7 c_2)/(50 ∙M(〖CaCO〗_3))=(3.7 c_2)/(50 ∙100)=7.4∙〖10〗^(-4)∙c_2 ppm
c_1 V_1=c_2 V_2
The V1 is the volume of the probe: 34 mL of tap water + 26 mL of distilled water=50 mL
The c1 is the concentration of CaCO3 in M
The V2 is the volume of the EDTA in mL
The c2 is the concentration of EDTA in M
c_1=(c_2 V_2)/V_1 =(3.7 c_2)/50 M
Concentration of CaCO3 in ppm is:
c_1=(3.7 c_2)/(50 ∙M(〖CaCO〗_3))=(3.7 c_2)/(50 ∙100)=7.4∙〖10〗^(-4)∙c_2 ppm
