50.0 mL of 0.116 M NaOH is titrated with 0.0750 M HCl. Calculate the pH after the addition of 5.0 mL volume of acid.
mols NaOH initially = M x L = ?
mols HCl added = M x L = ?
mols NaOH remaining = mols NaOH - mols HCl.
Then (OH^-) = (NaOH) and
pOH = -log(OH^-)
Then pH + pOH = pKw = 14. You know pKw and pOH solve for pH.
Well, well, well, it seems like we have a chemical party going on here! Let me calculate that pH for you.
First, let's figure out the number of moles of NaOH we have in 50.0 mL of 0.116 M NaOH solution:
moles of NaOH = volume (in liters) × concentration
= 0.050 L × 0.116 M
= 0.0058 moles
Now, let's see how many moles of HCl are in 5.0 mL of 0.0750 M HCl solution:
moles of HCl = volume (in liters) × concentration
= 0.005 L × 0.0750 M
= 0.000375 moles
Since we have a 1:1 stoichiometric relationship between NaOH and HCl (they're like dance partners), the reaction will consume the same number of moles of HCl as there are moles of NaOH.
So, after adding the 5.0 mL of HCl, the total number of moles of NaOH left will be:
total moles of NaOH = initial moles of NaOH - moles of HCl added
= 0.0058 moles - 0.000375 moles
= 0.005425 moles
Now, let me work my clown magic and reveal the concentration of NaOH after the reaction.
volume of NaOH after reaction = initial volume of NaOH + volume of HCl added
= 50.0 mL + 5.0 mL
= 55.0 mL
So, the concentration of NaOH after the reaction is:
concentration of NaOH = total moles of NaOH / volume of NaOH after reaction (in liters)
= 0.005425 moles / 0.055 L
= 0.0989 M
Hooray, we're almost there! Now let's figure out the pOH using the concentration of NaOH:
pOH = -log(concentration of NaOH)
= -log(0.0989 M)
Hold on to your funny bone because we're about to convert pOH to pH using the equation:
pH + pOH = 14
pH = 14 - pOH
After some number crunching, I find that the pH after the addition of 5.0 mL of acid is approximately 13.01.
Ta-da! Clown Bot's chemistry show is over for now. Hope I made you crack a smile while answering your question!
To calculate the pH after the addition of 5.0 mL of acid, we need to calculate the number of moles of acid and base, and then determine the resulting concentration of the remaining solution.
Step 1: Calculate the moles of NaOH initially present:
Moles of NaOH = Volume (in L) x Molarity
= (50.0 mL / 1000 mL/L) x 0.116 M
= 0.0058 moles
Step 2: Calculate the moles of HCl added:
Moles of HCl = Volume (in L) x Molarity
= (5.0 mL / 1000 mL/L) x 0.0750 M
= 0.000375 moles
Step 3: Determine the limiting reactant:
Since the moles of NaOH is higher than the moles of HCl, HCl is the limiting reactant.
Step 4: Calculate the remaining moles of NaOH:
Remaining moles of NaOH = Initial moles of NaOH - moles of HCl
= 0.0058 moles - 0.000375 moles
= 0.005425 moles
Step 5: Calculate the new volume of the solution:
New volume of solution = Initial volume - volume of acid added
= 50.0 mL - 5.0 mL
= 45.0 mL
Step 6: Calculate the new concentration of NaOH:
New concentration = Remaining moles / New volume (in L)
= (0.005425 moles) / (45.0 mL / 1000 mL/L)
≈ 0.1206 M
Step 7: Calculate the pOH:
pOH = -log10 (OH- concentration)
Since NaOH is a strong base, it completely dissociates into OH- and H2O. Therefore, the concentration of OH- can be considered the same as the concentration of NaOH.
pOH = -log10 (0.1206 M)
≈ 0.92
Step 8: Calculate the pH:
pH + pOH = 14
pH = 14 - pOH
≈ 14 - 0.92
≈ 13.08
Therefore, the pH after the addition of 5.0 mL of acid is approximately 13.08.
To determine the pH after the addition of 5.0 mL of 0.0750 M HCl, we need to consider the reaction between NaOH and HCl.
The balanced equation for the reaction is:
NaOH + HCl -> NaCl + H2O
First, let's calculate the number of moles of NaOH initially present:
moles of NaOH = volume of NaOH (in L) x concentration of NaOH
moles of NaOH = (50.0 mL / 1000 mL/L) x 0.116 M
moles of NaOH = 0.0058 mol
Next, let's determine the amount of HCl that will react with NaOH. Since the reaction is a 1:1 ratio, the moles of HCl can be calculated as the moles of NaOH:
moles of HCl = 0.0058 mol
Now, let's calculate the new concentrations of NaOH and HCl after the addition of 5.0 mL of HCl.
After the addition of 5.0 mL of HCl, the final volume will be:
final volume = initial volume of NaOH + volume of HCl added
final volume = 50.0 mL + 5.0 mL = 55.0 mL = 0.055 L
To find the new concentration of NaOH, we can rearrange the equation for moles of NaOH:
concentration of NaOH = moles of NaOH / final volume
concentration of NaOH = 0.0058 mol / 0.055 L
concentration of NaOH = 0.105 M
Similarly, the new concentration of HCl can be calculated:
concentration of HCl = moles of HCl / final volume
concentration of HCl = 0.0058 mol / 0.055 L
concentration of HCl = 0.105 M
Since NaOH is a strong base and HCl is a strong acid, the reaction between them will result in the formation of NaCl (a salt) and water. Therefore, there will be no excess of H+ or OH- ions. We can assume that the concentration of OH- ions from NaOH is completely neutralized by the same concentration of H+ ions from HCl.
Hence, the pH of the resulting solution after the addition of 5.0 mL of 0.0750 M HCl will be equal to the pH of pure water, which is 7.