when 70.mL of 3.0M Na2CO3 is added to 30.mL of 1.0M NaHCO3 the resulting concentration of Na+ is...?

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  • 1 month ago

    Supposing complete dissociation in both cases:

    Na2CO3 → 2 Na{+} + CO3{2-}

    NaHCO3 → Na{+} + H{+} + CO3{2-}

    (0.070 L) x (3.0 mol/L) x (2 mol Na{+} / 1 mol Na2CO3) =

    0.42 mol Na{+} from Na2CO3

    (0.030 L) x (1.0 mol/L) x (1 mol Na{+} / 1 mol NaHCO3) =

    0.030 mol Na{+} from NaHCO3

    Supposing additive volumes:

    (0.42 mol + 0.030 mol) / (0.070 L + 0.030 L) = 4.5 mol/L = 4.5 M Na{+}

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