Solid sodium carbonate (Na2CO3) is available in highly pure form and is often us
ID: 834909 • Letter: S
Question
Solid sodium carbonate (Na2CO3) is available in highly pure form and is often used to
determine the concentration of acidic solutions.
A lab technician needs to prepare 75.0 mL of a 0.425 M solution of Na2CO3 by dissolving
solid Na2CO3 in water. What mass of Na2CO3 should be used?
Na2CO3 reacts with HCl to produce H2CO3 and NaCl. If 45.0-mL of the 0.425 M solution
from part (a) required 28.3 mL of an HCl solution to reach the equivalence point, what was the
molarity of the HCl solution?
(a). 3.38 g Na2CO3 (b). 1.35 M HCl
Explanation / Answer
) MOLARITY=wt.of Na2CO3*1000/(M.WT OF Na2CO3*V(ml))
0.425=wt.of Na2CO3*1000/(106*75)
WT.OF Na2CO3=3.378 grams is required
2) Na2CO3 + 2 HCl--------------> 2 NaCl + H2CO3
according to dilution law
M1V1/n1= M2V2/n2
0.425*45/1=M2*28.3/2
M2=1.335M of HCl is required
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