H2SO4(aq) + NaOH(aq)-> Na2SO4(aq) + H2O(l)
Consider the unbalanced equation above. What is the concentration of the original sulfuric acid solution if 60.5 mL of the acid reacts with 18.0 mL of 0.110 M sodium hydroxide?
1. Balance the equation.
2. mols NaOH = M x L = ?
3. Convert mols NaOH to mols H2SO4.
4. M H2SO4 = moles H2SO4/L H2SO4.
To determine the concentration of the original sulfuric acid solution, we need to first balance the chemical equation:
H2SO4(aq) + 2NaOH(aq) -> Na2SO4(aq) + 2H2O(l)
Now let's calculate the amount of sodium hydroxide (NaOH) reacting with sulfuric acid (H2SO4). We have 18.0 mL of 0.110 M NaOH, which means we have:
(18.0 mL) x (0.110 mol/L) = 1.98 mmol of NaOH
According to the balanced equation, 2 moles of NaOH react with 1 mol of H2SO4. Therefore, the number of moles of H2SO4 reacting with NaOH is half the number of moles of NaOH:
1.98 mmol of NaOH / 2 = 0.99 mmol of H2SO4
Now we can calculate the concentration of the original sulfuric acid solution. We have 60.5 mL of the acid solution, so:
(0.99 mmol) / (60.5 mL) = 0.0163 M
Therefore, the concentration of the original sulfuric acid solution is approximately 0.0163 M.
To find the concentration of the original sulfuric acid solution, we can use the stoichiometry of the balanced equation.
First, let's balance the equation:
H2SO4(aq) + 2NaOH(aq) -> Na2SO4(aq) + 2H2O(l)
From the balanced equation, we can see that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide.
Next, let's convert the given volume of sodium hydroxide solution to moles:
Moles of NaOH = Volume (L) x Concentration (M)
Moles of NaOH = 0.018 L x 0.110 M
Moles of NaOH = 0.00198 mol
Since sulfuric acid reacts with twice the number of moles of sodium hydroxide, we can say that 0.00198 mol of sulfuric acid reacts.
Now, we can calculate the concentration of the original sulfuric acid solution.
Molarity = Moles of solute / Volume of solution (L)
Let's assume the volume of the original sulfuric acid solution is V L.
Molarity of sulfuric acid solution = 0.00198 mol / V L
We are given that 60.5 mL of the acid reacts, so we need to convert this to liters:
60.5 mL = 0.0605 L
Now, we can solve for the concentration:
0.00198 mol / 0.0605 L = 0.0326 M
Therefore, the concentration of the original sulfuric acid solution is 0.0326 M.