# How many grams of KOH are needed to neutralise 500mL of 2mol L^-1 H2SO4?

Oct 27, 2015

112.22 grams KOH

#### Explanation:

$\textcolor{g r e e n}{\text{Step 1:}}$

The first thing you need to do is write the chemical equation.

$K O H$ + ${H}_{2} S {O}_{4}$ = ${K}_{2} S {O}_{4}$ + ${H}_{2} O$ (unbalanced)

$\textcolor{g r e e n}{\text{Step 2:}}$

Balance the chemical equation. You need to do this because it's the only way that you can get the "conversion factor" you needed to solve the problem.

$\textcolor{red}{2} K O H$ + ${H}_{2} S {O}_{4}$ = ${K}_{2} S {O}_{4}$ + $\textcolor{red}{2} {H}_{2} O$ (balanced)

left side:

K = 1 x $\textcolor{red}{2}$ = 2
O = (1 x $\textcolor{red}{2}$) + 4 = 6
H = (1 x $\textcolor{red}{2}$) + 2 =4
S = 1

right side:

K = 2
O = 4 + (1 x $\textcolor{red}{2}$) = 2
H = 2 x $\textcolor{red}{2}$ = 4
S = 1

$\textcolor{g r e e n}{\text{Step 3:}}$

Convert the volume of ${H}_{2} S {O}_{4}$ into number of moles. How? By multiplying the given volume to its given molarity. Notice that I already converted the unit from milliliter (mL) to liter (L).

volume ${H}_{2} S {O}_{4}$ = 500 mL = 0.5 L
concentration of ${H}_{2} S {O}_{4}$ = 2M or $2 m o l \cdot {L}^{\text{-1}}$

$0.5 \cancel{L}$ ${H}_{2} S {O}_{4}$ x "2 moles"/(1 cancel "L") ${H}_{2} S {O}_{4}$ = 1 mole ${H}_{2} S {O}_{4}$

$\textcolor{g r e e n}{\text{Step 4:}}$

Now, going back to the balanced equation,

$\textcolor{red}{2} K O H$ + $\textcolor{red}{1} {H}_{2} S {O}_{4}$ = $\textcolor{red}{1} {K}_{2} S {O}_{4}$ + $\textcolor{red}{2} {H}_{2} O$

we know that

$\textcolor{red}{1}$ mole ${H}_{2} S {O}_{4}$ = $\textcolor{red}{2}$ moles $K O H$

Therefore, to convert the computed mole of ${H}_{2} S {O}_{4}$ into mole of $K O H$,

$1$ $\cancel{\text{mol}}$${H}_{2} S {O}_{4}$ x (2 "mol" KOH)/(1 cancel ("mol") H_2SO_4) = $2$ $\text{mol}$ $K O H$

$\textcolor{g r e e n}{\text{Step 5:}}$

Getting the atomic weights from the periodic table, we can compute the molar mass of $K O H$,

K = $39.10 \text{g/mol}$
O = $16.00 \text{g/mol}$
H = $1.01 \text{g/mol}$

molar mass of $K O H$ = $39.10 \text{g/mol}$ + $16.00 \text{g/mol}$ + $1.01 \text{g/mol}$ = $56.11 \text{g/mol}$

$\textcolor{g r e e n}{\text{Step 6:}}$

Converting the number of moles we got from $\textcolor{g r e e n}{\text{Step 4}}$ using the molar mass of $K O H$ obtained in $\textcolor{g r e e n}{\text{Step 5}}$,

$2$ $\cancel{\text{mol}}$$K O H$ x $\frac{56.11 g r a m s K O H}{1 \cancel{\text{mol}} K O H}$ = $\textcolor{red}{112.22 g K O H}$