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# When 50.0 millimeters of 3.0 - molar NaHCO3 is added to 50.0 millimeters of 1.0-molar Na2CO3 the resulting concentration of NA+ is?

When 50.0 millimeters of 3.0 - molar NaHCO3 is added to 50.0 millimeters of 1.0-molar Na2CO3 the resulting concentration of NA+ is?

I need to find millimeters or use empirical to find the concentration of NA+?

### 3 Answers

- Roger the MoleLv 76 months agoFavorite Answer
Supposing you meant "milliliters" instead of "millimeters", and supposing complete ionization for both salts:

NaHCO3 → Na{+} + HCO3{-}

Na2CO3 → 2 Na{+} + CO3{2-}

(50.0 mL) x (3.0 M NaHCO3) x (1 mol Na{+} / 1 mol NaHCO3) = 150. mmol Na{+} from NaHCO3

(50.0 mL) x (1.0 M Na2CO3) x (2 mol Na{+} / 1 mol Na2CO3) = 100. mmol Na{+} from Na2CO3

(150. mmol Na{+} + 100. mmol Na{+}) / (50.0 mL + 50.0 mL) = 2.5 mmol/mL = 2.5 mol/L Na{+}

- Trevor HLv 76 months ago
If we overlook the errors in the question - you rally must pay attention to what you submit!:

Mol NaHCO3 in 50mL of 3.0M solution = 50.0mL / 1000mL/L * 3.0mol/L = 0.15 mol NaHCO3

Because 1 mol NaHCO3 contains 1 mol Na+ - you have 0.15 mol Na+

Mol Na2CO3 in 50.0mL of 1.0M solution = 50mL/1000mL/L * 1.0mol/L = 0.05 mol Na2CO3

Because 1 mol na2CO3 contains 2 mol Na+, you have 0.05*2 = 0.10 mol Na+

Total mol Na+ = 0.15mol + 0.05 mol = 0.20mol Na+

Volume of solution =50.0mL + 50.0mL = 100.0mL = 0.100L

Molar concentration of Na+ in solution = 0.20mol / 0.100L = 1.0M

wouldn't it be .25 mol since there's 2 mol of Na{+} and you add 0.15 + .10 mol instead of 0.05 mol?

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- billrussell42Lv 76 months ago
millimeters (mm) are a measure of length. I suspect you mean milliliters (mL) ?

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Another way to check is to notice that the final mixture must have a concentration somewhere between the concentrations of the two given solutions. 0.0025 mol/L is not between 3.0 M and 1.0 M, so it cannot be the answer.