Question

If 10.0 mL of 1.0 M HBr is added to 1.00 L of the solution in question 3, what is the resulting solution’s pH?

Question 3 and results****

What is the pH of a solution with a HBrO concentration of 0.10 M and a NaBrO concentration of 0.20 M?

Br₂(l) + H₂O(l) ↔ HOBr(aq) + HBr (aq)
HOBr(l) + NaOH(l) ↔ NaOBr(aq) + H₂O (aq)

by using hendersen-hasselbalch equation we get,
pK = pH + log {[conj.base]/[conj.acid]}

{by standard we know pK =14}

14 = pH + log {[0.20]/[0.10]}
14 = pH + 0.3010

pH = 13.699

Answer 1

Br₂(l) + H₂O(l) ↔ HOBr(aq) + HBr (aq)
HOBr(l) + NaOH(l) ↔ NaOBr(aq) + H₂O (aq)

by using hendersen-hasselbalch equation we get,
pK = pH + log {[conj.base]/[conj.acid]}

{by standard we know pK =14}

14 = pH + log {[0.20]/[0.10]}
14 = pH + 0.3010

pH = 13.699

10.0 mL of 1.0 M HBr is added to 1.00 L of the solution in question 3

no of moles of HBr = molarity *volume in L

                              = 1*0.01 = 0.01moles of HBr

no of moles of HBrO = 0.1+0.01 = 0.11moles

no of moles of NaOBr = 0.2 + 0.01 = 0.21 moles

PH = PKa + log[NaOBr]/[HOBr]

     = 8.63+ log0.21/0.11

      =8.63 + 0.28 = 8.91