# 115 mL of a 2.35 M potassium fluoride solution is diluted with 1.28L of water. What is the new concentration in molarity?

Aug 19, 2016

$\left[K F\right]$ $=$ $0.194 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

$\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of solution}$

$\frac{0.115 \cdot L \times 2.35 \cdot m o l \cdot {L}^{-} 1}{0.115 \cdot L + 1.28 \cdot L}$

$=$ $\frac{0.270 \cdot m o l}{1.395 \cdot L}$ $=$ $0.194 \cdot m o l \cdot {L}^{-} 1$.

Please note that the wording of the problem said that the initial $0.115 \cdot L$ volume was diluted with $1.28 \cdot L$ FRESH SOLVENT. Had it said $\text{diluted to 1.28} \cdot L$, the volume would change accordingly. Do you see why I make the distinction?

Such a solution would be slightly basic. Why?