# 0.650 L of 0.400 M H_2SO_4 is mixed with 0.600 L of 0.280 M KOH. What concentration of sulfuric acid remains after neutralization?

Mar 21, 2018

$\text{0.141 M}$.

#### Explanation:

Here is our balanced neutralisation equation:

${H}_{2} S {O}_{4} \left(a q\right) + 2 K O H \left(a q\right) \to 2 {H}_{2} O \left(l\right) + {K}_{2} S {O}_{4} \left(a q\right)$

From this, we know that for every $1$ mole of ${H}_{2} S {O}_{4}$ that reacts, $2$ moles of $K O H$ reacts. In other words, the mole ratio of ${H}_{2} S {O}_{4}$ to $K O H$ is $1 : 2$.
To find the concentration of ${H}_{2} S {O}_{4}$ left after neutralisation, we just need to find how much it's in excess.

$\text{0.650 L}$ of $\text{0.400 M}$ ${H}_{2} S {O}_{4}$ is $\text{0.650 L" xx "0.400 mol/L} = 0.260$ moles of ${H}_{2} S {O}_{4}$.
$\text{0.600 L}$ of $\text{0.280 M}$ $K O H$ is $\text{0.600 L" xx "0.280 mol/L} = 0.168$ moles of $K O H$.

We know that the mole ratio of $K O H$ to ${H}_{2} S {O}_{4}$ is $1 : 2$, so, to fully react with the $0.168$ moles of $K O H$, $\frac{0.168}{2} = 0.084$ moles of ${H}_{2} S {O}_{4}$ is needed.

We have $0.260$ moles.

This means that only $0.084$ moles out of $0.260$ moles of ${H}_{2} S {O}_{4}$ react. $0.260 - 0.084 = 0.176$ moles of ${H}_{2} S {O}_{4}$ don't react at all.

Now that we know how much ${H}_{2} S {O}_{4}$ is left sitting there in the solution ($0.176$), we can use that to calculate its concentration in molarity.

$\text{molarity" = "moles" / "volume (L)}$

The number of moles is $0.176$, and the volume is $\text{0.650 L + 0.600 L = 1.25 L}$. Plugging these values into the equation, we get:

$\text{molarity" = "0.176 moles" / "1.25 L" = "0.141 M}$