Question

What volume of 6 M HCI and 2 M HCI should be mixed to get litre of 3M HCI?

Answer 1

0.25 L of 6 M HCl and 0.75 L of 2 M HCl

Answer 2

To determine the volumes of 6 M HCl and 2 M HCl needed to obtain 1 litre of 3 M HCl, we use the principle of conservation of moles. When solutions are mixed, the total number of moles of the solute remains constant.

Let $V_1$ be the volume (in litres) of 6 M HCl and $V_2$ be the volume (in litres) of 2 M HCl.
The total volume of the final solution is 1 litre.
So, we have the first equation:
$V_1 + V_2 = 1 \text{ L}$ (Equation 1)

The total moles of HCl in the final mixture must be equal to the sum of moles from the initial solutions. The number of moles is given by Molarity $\times$ Volume.
Moles from 6 M HCl = $6 \times V_1$
Moles from 2 M HCl = $2 \times V_2$
Moles in final 3 M HCl solution = $3 \times 1 \text{ L} = 3 \text{ moles}$

Using the conservation of moles:
$M_1V_1 + M_2V_2 = M_{final}V_{final}$
$6V_1 + 2V_2 = 3 \times 1$
$6V_1 + 2V_2 = 3$ (Equation 2)

Now we have a system of two linear equations:

1. $V_1 + V_2 = 1$
2. $6V_1 + 2V_2 = 3$

From Equation 1, express $V_2$ in terms of $V_1$:
$V_2 = 1 - V_1$

Substitute this expression for $V_2$ into Equation 2:
$6V_1 + 2(1 - V_1) = 3$
$6V_1 + 2 - 2V_1 = 3$
$4V_1 + 2 = 3$
$4V_1 = 3 - 2$
$4V_1 = 1$
$V_1 = \frac{1}{4} \text{ L}$
$V_1 = 0.25 \text{ L}$

Now, substitute the value of $V_1$ back into Equation 1 to find $V_2$:
$V_2 = 1 - V_1$
$V_2 = 1 - 0.25$
$V_2 = 0.75 \text{ L}$

So, 0.25 litres of 6 M HCl and 0.75 litres of 2 M HCl should be mixed.
In millilitres:
$V_1 = 0.25 \text{ L} = 250 \text{ mL}$
$V_2 = 0.75 \text{ L} = 750 \text{ mL}$