If 100mL of a 0.10M NaOH solution is added to 75mL of a 0.15 HCl solution, then what is the pH of the resultant solution?
Write the equation.
HCl + NaOH ==> NaCl + H2O
mols NaOH = M*L = 0.10 M x 0.1 L = ??
mols HCl = M*L = 0.15 M x 0.075 L = ??
See which is in excess, how much is left over after reacting, then pH = - log(H^+). Post your work if you get stuck.
I don't understand where the H+ comes from, I mean I know it comes from the acid, but don't I need an equilibrium constant K to solve for the H+?
No. Since the product is NaCl and water and the NaCl doesn't hydrolyze, the H+ or OH- comes from the excess of HCl OR NaOH. Look at the mols HCl and the mols NaOH. Calculate each according to my last post. You will see one is larger than the other. So subtract them to find the difference and that will be the H+ if it is the HCl in excess OR it will be the OH- if it is the NaOH in excess. Both HCl AND NaOH are strong acids and bases so neither has a Ka or Kb.
A note that when you subtract the difference is the MOLS of excess reagent. The concentration is mols/L to calculate H+ or OH- so mols/0.175 L is the concentration.
To determine the pH of the resultant solution, we need to consider the reaction that occurs when NaOH reacts with HCl. The balanced chemical equation for the reaction is:
NaOH + HCl → NaCl + H2O
This reaction between the strong base (NaOH) and strong acid (HCl) forms a neutral salt (NaCl) and produces water (H2O). Since the reaction goes to completion, we can assume that all the NaOH and HCl react completely, leaving no excess of either reactant.
To find the pH of the resultant solution, we need to calculate the concentration of H+ ions (protons) in the solution. Since the reaction fully neutralizes the acid and base, the concentration of H+ and OH- ions in the final solution will be equal. This means that the concentration of H+ ions in the final solution is determined by the excess of the stronger acid or stronger base.
In this case, the HCl solution is the stronger acid compared to NaOH. We need to determine how much of the acid solution is left after the reaction takes place.
First, let's calculate the number of moles of HCl and NaOH using the given concentrations and volumes:
Moles of HCl = Concentration of HCl × Volume of HCl solution
= 0.15 M × 75 mL
= 0.01125 moles
Moles of NaOH = Concentration of NaOH × Volume of NaOH solution
= 0.10 M × 100 mL
= 0.01 moles
Now, we need to compare the number of moles of HCl and NaOH to determine which reactant is in excess.
Since the stoichiometry of the balanced chemical equation is 1:1, we can see that HCl is in excess because it has a higher number of moles (0.01125 moles) compared to NaOH (0.01 moles).
Since all the excess HCl has reacted, we can find the concentration of H+ ions in the resultant solution by only considering the remaining NaOH.
Therefore, the concentration of H+ ions can be calculated by dividing the number of moles of NaOH by the total volume of the resultant solution:
Concentration of H+ ions = Moles of NaOH / Total volume of the resultant solution
The total volume of the resultant solution is the sum of the volumes of the NaOH and HCl solutions:
Total volume of resultant solution = Volume of NaOH solution + Volume of HCl solution
= 100 mL + 75 mL
= 175 mL
Concentration of H+ ions = 0.01 moles / 175 mL
Now, to find the pH, we need to calculate the negative logarithm (base 10) of the H+ ion concentration using the formula:
pH = -log10 [H+]
Calculate the value by substituting the concentration of H+ ions into the formula:
pH = -log10 (0.01 moles / 175 mL)
After performing the calculations, you will obtain the pH value of the resultant solution.