# A second reaction mixture was made up in the following way: 5 mL 4.0 M acetone + 20 Ml 1.0 M HCL...

## Question:

A second reaction mixture was made up in the following way:

5 mL 4.0 M acetone + 20 Ml 1.0 M HCL + 10 mL 0.0050 M I 2 + 15 mL H20

What were the initial concentrations of acetone, H+ ion, and I2 in the reaction mixture?

## Molarity

Molarity is a concentration unit defined as the moles of solute divide by the liter of solution. Molarity is also an intensive property in that it is not affected by the size of the sample.

{eq}[H^+] = 0.4\;M{/eq}

{eq}[I_2] = 0.001\;M{/eq}

First, we need to find out the moles of {eq}H^+{/eq} and {eq}I_2{/eq}. To do this, we use the volume and concentration of HCl and {eq}I_2{/eq} given in the problem.

Moles of {eq}H^+{/eq} can be found as:

{eq}(0.020\;L)(1.0\;M) = 0.020\;moles \;H^+{/eq}

Moles of {eq}I_2{/eq} can be found as:

{eq}(0.010\;L)(0.0050\;M) = 0.00005\;moles\;I_2{/eq}

The total volume of the mixture is:

{eq}(5\;mL + 20\;mL + 10\;mL + 15\;mL) \times \frac {1\;L}{1000\;mL}= 0.05\;L{/eq}

So the concentration of {eq}H^+{/eq} is:

{eq}\frac {0.020\;moles \;H^+}{0.05\;L} = \boxed {0.4\;M}{/eq}

And the concentration of {eq}I_2{/eq} is:

{eq}\frac {0.00005\;moles\;I_2}{0.05\;L} = \boxed {0.001\;M}{/eq}. 