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Marat540 [252]
3 years ago
10

How many ml of 0.01-m naoh will be required to titrate 0.061-g of khp to it's endpoint?

Chemistry
1 answer:
hjlf3 years ago
6 0
The volume of NaOH needed for the titration is calculated by equating the number of moles of KHP and NAOH.

(1) number of moles of KHP:
         n = mass of KHP / molar mass of KHP
Substituting,
         n = 0.061 g / (204.22 g/mol) = 2.987 x 10⁻⁴ moles KHP

(2) number of moles of NaOH
         n = (volume NaOH in L)(molarity of NaOH)
          n = V x 0.01

Equating,
       0.01V = 2.987 x 10⁻⁴

The value of V from the equation is 0.0299L of NaOH. 

Converting the value to mL will give us the answer of 29.87 mL. 
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To find the fraction of acetic acid that is in the dissociated form (f), we apply the following equation (Ka for acetic acid is 1.76*10⁻⁵). This equation comes from solving the equation of the equilibrium constant for the dissociated fraction of HAc:

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Explanation:

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