A 25.00 mL sample of 0.320 M KOH is titrated with 0.750 M HBr at 25 °C. Required:
a. Calculate the initial pH before any titrant is added.
b. What is the initial pH before any titrant was added?
c. What is the pH of the solution after 5.00mL of the titrant was added?

Given that :

Reaction equation ;

A) and B)  Determine the Initial pH before addition of titrant

First step : compute the value of pOH

pOH = - log ( [OH⁻])

        = -log ( 0.320 ) = 0.50

∴ Initial pH before the titrant is added = 14 - pOH

                                                               = 14 - 0.5 = 13.5

C ) Determine the pH of solution after 5 mL titrant is added

First step : determine the reacting moles of each substance

n kOH = 0.025 mL * 0.320 mol/L = 0.00800 mol

n HBr = 0.005 * 0.750 mol/L = 0.00375 mol

next step : compute the value of the concentration of OH⁻ in final solution

n kOH ( remaining ) = 0.00800 - 0.00375 = 0.00425 mol

final solution = 25.00 mL + 5.00 mL = 30.00 mL

∴ The value of the concentration of [ OH⁻] in 30.00 mL

  = [ OH⁻] = ( 0.0042 mol / 0.0300 mL )

                = 0.142 M

 hence pOH = - log (0.142) = 0.849

Final step; The pH of the solution after 5 mL of titrant is added

= 14 - pOH

= 14 - 0.849 = 13.15

Hence we can conclude that the initial pH before titrant was added is 13.50 and the pH after is as listed above.

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the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying