Calculate the [H+] of a solution obtained by mixing 1 L of hydrochloric acid 1.0 M with 1 litre of sodium hydroxide 0.990 M.
mols HCl = M x L = ?
mols NaOH = M x L = ?
Subtract. Excess mols = NaOH, calculate H^+.
To calculate the [H+] (concentration of hydrogen ions) of the resulting solution, we can use the concept of neutralization.
Neutralization occurs when an acid reacts with a base, producing water and a salt. In this case, hydrochloric acid (HCl) is a strong acid, and sodium hydroxide (NaOH) is a strong base.
The balanced chemical equation for the neutralization reaction is:
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
According to the stoichiometry of the equation, one mole of HCl reacts with one mole of NaOH to produce one mole of water.
Given that 1 L of hydrochloric acid 1.0 M and 1 L of sodium hydroxide 0.990 M are mixed together, we can determine the moles of HCl and NaOH present:
Moles of HCl = 1.0 mol/L × 1 L = 1.0 mol
Moles of NaOH = 0.990 mol/L × 1 L = 0.990 mol
Since the stoichiometry in the balanced equation is 1:1 between HCl and NaOH, the limiting reactant is NaOH. This means that all of the NaOH will react with the HCl, leaving some HCl unreacted.
To find the excess HCl, we can subtract the moles of NaOH from the initial moles of HCl:
Excess moles of HCl = Moles of HCl - Moles of NaOH
= 1.0 mol - 0.990 mol
= 0.010 mol
The remaining moles of HCl will contribute to the [H+] in the resulting solution.
The total volume of the solution is 1 L + 1 L = 2 L.
To calculate the [H+], we divide the excess moles of HCl by the total volume of the solution in liters:
[H+] = (Excess moles of HCl) / (Total Volume)
= 0.010 mol / 2 L
= 0.005 M
Therefore, the [H+] of the resulting solution is 0.005 M.
To calculate the [H+] (concentration of hydrogen ions) in the resulting solution obtained by mixing hydrochloric acid and sodium hydroxide, we need to consider the stoichiometry of the reaction between these two compounds.
The balanced chemical equation for the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) is as follows:
HCl + NaOH -> NaCl + H2O
From the equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of water (H2O). Therefore, the concentration of hydrogen ions in the final solution will depend on the amount of excess HCl or NaOH.
Step 1: Determine the limiting reagent
To determine which reactant is the limiting reagent, we compare the number of moles of HCl and NaOH.
1 L of 1.0 M HCl solution contains:
1.0 mole/L * 1 L = 1.0 moles of HCl
1 L of 0.990 M NaOH solution contains:
0.990 mole/L * 1 L = 0.990 moles of NaOH
Since we have an equal number of moles (1.0 moles of HCl and 0.990 moles of NaOH), we have an equal number of moles of the reactants. Therefore, neither HCl nor NaOH is in excess, and both will be completely consumed in the reaction.
Step 2: Calculate the number of moles of water produced
Since the reaction is balanced on a 1:1 ratio, the number of moles of water produced will be the same as the number of moles of HCl or NaOH used in the reaction.
Therefore, 1.0 moles of water will be produced.
Step 3: Calculate the final volume of the solution
When HCl and NaOH react, they form water and sodium chloride (NaCl).
Since the unreacted HCl and NaOH react to form water, the final volume of the solution will remain at 1 L.
Step 4: Calculate the concentration of hydrogen ions ([H+])
To calculate the concentration of hydrogen ions, we divide the number of moles of HCl by the final volume of the solution.
[H+] = (moles of HCl) / (volume of solution)
[H+] = 1.0 moles / 1 L
[H+] = 1.0 M
Therefore, the concentration of hydrogen ions in the resulting solution obtained by mixing 1 L of hydrochloric acid 1.0 M with 1 liter of sodium hydroxide 0.990 M is 1.0 M.