Calculate the pH of the resulting solution if 25.0 mL of 0.250 M HCl(aq) is added to
(a) 35.0 mL of 0.250 M NaOH(aq).
(b) 15.0 mL of 0.350 M NaOH(aq).
Some one please help
I found the answer to a however i keep doing something wrong for b
To calculate the pH of the resulting solution after mixing an acid and a base, you need to determine the concentration of the H+ ions in the solution. This can be done by considering the stoichiometry of the acid-base reaction and the balanced chemical equation.
(a) 35.0 mL of 0.250 M NaOH(aq):
Step 1: Write and balance the chemical equation for the reaction.
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Step 2: Determine the limiting reactant.
In this case, HCl and NaOH are in equimolar amounts, so neither is limiting.
Step 3: Calculate the total moles of hydroxide ions from NaOH:
moles NaOH = volume (L) × concentration (M)
moles NaOH = 0.035 L × 0.250 M = 0.00875 mol NaOH
Step 4: Calculate the total moles of hydrogen ions from HCl:
moles HCl = volume (L) × concentration (M)
moles HCl = 0.025 L × 0.250 M = 0.00625 mol HCl
Step 5: Determine the excess reactant and calculate the moles of excess reactant left after the reaction.
Since neither HCl nor NaOH is limiting, we need to determine which reactant produces the most moles of products. In this case, the limiting reactant is HCl, and NaOH is in excess.
moles of excess NaOH after reaction = moles NaOH initially - moles NaOH used
moles of excess NaOH = 0.00875 mol NaOH - 0.00625 mol NaOH = 0.0025 mol NaOH
Step 6: Determine the concentration of hydroxide ions (OH-) in the solution.
concentration of OH- = moles of excess NaOH / total volume of solution (L)
total volume of solution = volume of HCl + volume of NaOH
total volume of solution = 0.025 L + 0.035 L = 0.06 L
concentration of OH- = 0.0025 mol NaOH / 0.06 L = 0.0417 M OH-
Step 7: Convert the concentration of hydroxide ions to the concentration of hydrogen ions.
Since water dissociates to produce equal amounts of hydrogen ions (H+) and hydroxide ions (OH-), the concentration of H+ ions is the same as the OH- concentration.
concentration of H+ = 0.0417 M
Step 8: Calculate the pH of the solution.
pH = -log[H+]
pH = -log(0.0417)
pH ≈ 1.38
Therefore, the pH of the resulting solution is approximately 1.38.
(b) 15.0 mL of 0.350 M NaOH(aq):
Steps 1-5 are the same as for the previous calculation, so we will start from Step 6.
Step 6: Determine the concentration of hydroxide ions (OH-) in the solution.
You already calculated the moles of excess NaOH in Step 5.
concentration of OH- = moles of excess NaOH / total volume of solution (L)
total volume of solution = volume of HCl + volume of NaOH
total volume of solution = 0.025 L + 0.015 L = 0.04 L
concentration of OH- = 0.0025 mol NaOH / 0.04 L = 0.0625 M OH-
Step 7: Convert the concentration of hydroxide ions to the concentration of hydrogen ions.
As mentioned before, the concentration of H+ ions is the same as the OH- concentration.
concentration of H+ = 0.0625 M
Step 8: Calculate the pH of the solution.
pH = -log[H+]
pH = -log(0.0625)
pH ≈ 1.20
Therefore, the pH of the resulting solution is approximately 1.20.
My guess is that you didn't take into account the total volume.
millimols HCl=mL x M = 25 x 0.250 = 6.25
mmols NaOH = 15 x 0.35 = 5.25
NaOH + HCl ==> H2O + NaCl
mmols HCl = 6.25
mmols NaoH = 5.25
difference is excess HCl by 1.00 mmol.
So you have 1.00 mmols HCl in a total volume of 40 mL.
M HCl = mmols/mL = ? and convert H^+ to pH.
Post your work if you still don't get the right answer.