Consider the titration of a 35.0-mL sample of 0.175 M HBr with 0.200 M KOH. Determine the pH after adding 5.0 mL of base beyond the equivalence point.

Determine where the equivalence point is. Note the volume.

mmoles OH^- added in excess is 5.0 mL x M = ??
total volume = mL to arrive at the equivalence point + 5.0 from the base
mmoles/total mL = M of OH^-
pOH = -log(OH^-)
and pH + pOH = pKw = 14. Solve for pH.

To determine the pH after adding 5.0 mL of base beyond the equivalence point, we need to go through the steps of the titration and understand the reaction that occurs.

Step 1: Write the balanced chemical equation for the reaction:
HBr + KOH → KBr + H2O

Step 2: Calculate the moles of acid and base used:

Moles of HBr = Volume (in L) x Concentration
= 0.035 L x 0.175 mol/L
= 0.006125 mol

Moles of KOH = Volume (in L) x Concentration
= 0.035 L x 0.200 mol/L
= 0.007 mol

Based on the balanced equation, we can see that 1 mole of KOH reacts with 1 mole of HBr to form 1 mole of water. This means that the moles of KOH used are in excess, while the moles of HBr used will react completely.

Step 3: Determine the moles of KOH left after the reaction:

Moles of KOH left = Moles of KOH used - Moles of HBr used
= 0.007 mol - 0.006125 mol
= 0.000875 mol

Step 4: Calculate the resulting concentration of KOH:

Volume of KOH left = Volume added - Volume reacted
= 0.005 L (since 5.0 mL is added)

Concentration of KOH = Moles of KOH left / Volume of KOH left
= 0.000875 mol / 0.005 L
= 0.175 M

Step 5: Calculate the concentration of hydroxide ions (OH-) in the solution:

Since KOH is a strong base, it fully dissociates in water, so the concentration of OH- ions is equal to the concentration of KOH:

Concentration of OH- = 0.175 M

Step 6: Calculate the pOH (power of hydroxide ion concentration):

pOH = -log10 (OH- concentration)

pOH = -log10 (0.175)
= 0.755

Step 7: Calculate the pH:

pH = 14 - pOH
= 14 - 0.755
= 13.245

Therefore, the pH after adding 5.0 mL of base beyond the equivalence point is approximately 13.245.

To determine the pH after adding 5.0 mL of base beyond the equivalence point, we need to consider the reaction between HBr and KOH and calculate the amount of excess OH^- ions remaining in the solution.

First, let's calculate the amount of HBr in the initial 35.0 mL solution:

moles of HBr = concentration of HBr * volume of HBr solution
= 0.175 M * 0.0350 L
= 0.006125 moles

Since KOH and HBr react in a 1:1 ratio, the amount of KOH required to neutralize the HBr completely is also 0.006125 moles.

Next, let's calculate the amount of the base added beyond the equivalence point:

amount of base added beyond equivalence point = 0.200 M * 0.0050 L
= 0.001 moles

Since we added 0.001 moles of KOH beyond the equivalence point and the reaction is complete, all of this KOH will react with water to form OH^- ions.

Now, let's calculate the total amount of OH^- ions present in the solution:

total moles of OH^- = moles of OH^- from excess KOH + moles of OH^- from water

The moles of OH^- from water can be calculated using the formula:

moles of OH^- from water = moles of added base beyond equivalence point

moles of OH^- from water = 0.001 moles

Since we added 0.001 moles of KOH beyond the equivalence point, the total moles of OH^- ions will be:

total moles of OH^- = 0.001 moles + 0.001 moles
= 0.002 moles

Now, let's calculate the concentration of OH^- ions:

total concentration of OH^- = total moles of OH^- / total volume of solution

The total volume of the solution is the sum of the initial volume of HBr and the volume of base added beyond the equivalence point:

total volume of solution = volume of HBr solution + volume of base added beyond equivalence point
= 35.0 mL + 5.0 mL
= 40.0 mL (or 0.0400 L)

total concentration of OH^- = 0.002 moles / 0.0400 L
= 0.050 M

Finally, we can calculate the pOH and then convert it to pH:

pOH = -log10(total concentration of OH^-)
= -log10(0.050)
= 1.30

pH = 14 - pOH
= 14 - 1.30
= 12.70

Therefore, the pH after adding 5.0 mL of base beyond the equivalence point is approximately 12.70.