Answer:
To solve this problem, we need to calculate the number of moles of NaOH and HCl that are added to the water, and then determine how these react to form a new solution.
First, let's calculate the number of moles of NaOH that are added to the water:
moles NaOH = Molarity x Volume (in liters)
moles NaOH = 0.100 mol/L x 0.00400 L
moles NaOH = 0.000400 mol
Next, let's calculate the number of moles of HCl that are added to the solution:
moles HCl = Molarity x Volume (in liters)
moles HCl = 0.800 mol/L x 0.01900 L
moles HCl = 0.0152 mol
Now we need to determine how these react with each other. NaOH and HCl react in a 1:1 ratio to form NaCl (sodium chloride) and water:
NaOH + HCl → NaCl + H2O
Since we have an excess of HCl, all of the NaOH will be used up in the reaction. Therefore, the moles of NaCl formed will be equal to the moles of NaOH added. The remaining HCl will determine the pH of the resulting solution.
moles NaCl = 0.000400 mol
The total volume of the resulting solution is:
volume = 50.00 mL + 4.00 mL + 19.00 mL
volume = 0.07300 L
The concentration of HCl in the resulting solution is:
concentration HCl = moles HCl / volume
concentration HCl = 0.0152 mol / 0.07300 L
concentration HCl = 0.208 M
To find the pH, we can use the formula:
pH = -log[H+]
where [H+] is the concentration of hydrogen ions in the solution. Since HCl is a strong acid, it completely dissociates in water to form H+ and Cl- ions. Therefore, the concentration of H+ ions in the solution is equal to the concentration of HCl.
pH = -log(0.208)
pH = 0.68
Therefore, the pH of the resulting solution is 0.68.