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What is the final $[{Na}^{+}]$ in a solution prepared by mixing 70.0 mL of $3.00 M {Na}_{2} {SO}_{4}$ with 30.0 mL of $1.00 M NaCl$ ?

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2.40 M

2.00 M

4.50 M

4.00 M

answer is D.

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Detailed Solution

$= \frac{(70 \times 3) \times 10^{- 3} \times 2 + (30 \times 1) \times 10^{- 3}}{(70 + 30) \times 10^{- 3}}$

$= \frac{420 + 30}{100} = \frac{400}{100} = 4.5$

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