Answer: 4.60
Explanation: To solve this problem, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa of the weak acid and the concentrations of the weak acid and its conjugate base:
pH = pKa + log([A^-]/[HA])
where [A^-] is the concentration of the conjugate base and [HA] is the concentration of the weak acid.
First, we need to calculate the initial concentrations of HClO and NaClO in the buffer solution:
[HClO] = 0.175 M
[NaClO] = 0.150 M
Next, we need to calculate the moles of HClO and NaClO in the buffer solution:
moles of HClO = 0.175 M x 0.100 L = 0.0175 moles
moles of NaClO = 0.150 M x 0.100 L = 0.0150 moles
Now we can use the stoichiometry of the reaction between HClO and HBr to calculate the change in concentration of HClO and NaClO after the addition of 150.0 mg of HBr:
HClO + HBr → H2O + Cl- + Br-
150.0 mg of HBr is equal to 0.1500 g. Assuming 100% dissociation of HBr, the moles of HBr added to the solution is:
moles of HBr = 0.1500 g / 80.91 g/mol = 0.001852 moles
From the balanced equation above, we can see that one mole of HBr reacts with one mole of HClO. Therefore, the change in concentration of HClO is equal to the moles of HBr added:
Δ[HClO] = -0.001852 moles
Similarly, we can see that one mole of HBr reacts with one mole of NaClO to form NaBr. Therefore, the change in concentration of NaClO is also equal to the moles of HBr added:
Δ[NaClO] = -0.001852 moles
Using the moles of HClO and NaClO in the buffer solution and the changes in concentration due to the addition of HBr, we can calculate the new concentrations of HClO and NaClO in the buffer solution:
[HClO] = (0.0175 - 0.001852) / 0.100 = 0.156 M
[NaClO] = (0.0150 - 0.001852) / 0.100 = 0.133 M
Now we can plug these concentrations into the Henderson-Hasselbalch equation to calculate the new pH of the buffer solution:
pH = pKa + log([A^-]/[HA])
pH = -log(2.9x10^-8) + log(0.133/0.156)
pH = 4.75 + (-0.154)
pH = 4.60
Therefore, the pH after the addition of 150.0 mg of HBr is 4.60.