Answer :
Answer:
The answer is "12.06"
Explanation:
Given:
[tex]M(HBrO) = 0.3\ M\\\\V(HBrO) = 20 \ mL\\\\M(KOH) = 0.15 \ M\\\\V(KOH) = 64 \ mL[/tex]
[tex]\to mol(HBrO) = M(HBrO) \times V(HBrO) = 0.3 M \times 20 mL = 6 \ mmol\\\\\to mol(KOH) = M(KOH) \times V(KOH)= 0.15 M \times 64 mL = 9.6 mmol[/tex]
6 mmol of both will react
excess KOH remaining[tex]= 3.15 \ mmol[/tex]
Volume[tex]= 20 + 64 = 84 \ mL[/tex]
[tex][OH^{-}] = \frac{ 9.6 \ mmol}{84\ mL} = 0.01142\ M[/tex]
use:
[tex]pOH = -\log [OH^-][/tex]
[tex]= -\log (1.142\times 10^{-2})\\\\= 1.94[/tex]
use:
[tex]PH = 14 - pOH[/tex]
[tex]= 14 - 1.94\\\\= 12.06[/tex]
The pH of the resulting solution is 12.63.
The equation of the reaction is;
HBrO(aq) + KOH(aq) ------> KOBr(aq) + H2O(l)
Number of moles of KOH= 64/1000 × 0.150 M = 0.0096 moles
Number of moles of HBrO = 20/1000 × 0.300 M = 0.0060 moles
Number of moles of excess base = 0.0096 moles - 0.0060 moles = 0.0036 moles
Total volume of solution = 64.0 mL + 20.0 mL = 84 mL = 0.084 L
Molarity of excess base = 0.0036 moles/0.084 L = 0.043 M
pOH = -log[OH-]
pOH = -log[ 0.043 M]
pOH = 1.37
pH + pOH = 14
pH = 14 -pOH
pH = 14 - 1.37
pH = 12.63
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