what is the pH of the solution formed when 25 mL of 0.173 M NaOH is added to 35 mL of 0.342 M HCl?
See above.
To find the pH of the solution, we need to determine the concentration of the resulting solution after the NaOH and HCl have reacted.
First, let's calculate the number of moles of NaOH and HCl.
Moles of NaOH = volume (in liters) × concentration = 0.025 L × 0.173 mol/L = 0.004325 mol
Moles of HCl = volume (in liters) × concentration = 0.035 L × 0.342 mol/L = 0.01197 mol
Since NaOH and HCl react in a 1:1 ratio, the limiting reactant is the one with fewer moles, in this case, NaOH. This means that all of the NaOH will react with an equal number of moles of HCl.
Therefore, the remaining moles of HCl after the reaction will be:
Remaining moles of HCl = Initial moles of HCl - reacted moles of NaOH = 0.01197 mol - 0.004325 mol = 0.007645 mol
Next, we need to calculate the combined volume of the NaOH and HCl solutions.
Combined volume = volume of NaOH + volume of HCl = 25 mL + 35 mL = 60 mL = 0.060 L
To find the concentration of the resulting solution, we divide the moles of HCl by the combined volume.
Concentration of the resulting solution = moles of HCl / combined volume = 0.007645 mol / 0.060 L ≈ 0.1274 M
Now that we have the concentration of the resulting solution, we can find the pH. The pH of a solution is given by the equation:
pH = -log[H+]
Since HCl is a strong acid, it completely dissociates in water, yielding H+ ions. Therefore, the concentration of H+ ions in the resulting solution is equal to the concentration of the HCl solution, which is 0.1274 M.
pH = -log(0.1274) ≈ 0.893
Therefore, the pH of the solution formed when 25 mL of 0.173 M NaOH is added to 35 mL of 0.342 M HCl is approximately 0.893.