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Wewaii [24]
3 years ago
5

You have 125.0 mL of a solution of H3PO4, bu you don't know its concentration. if you titrate the solution with a 4.56-M solutio

n of NaOH and reach the endpoint when 134.1 mL of the base are added, what is the concentration of the acid
Chemistry
1 answer:
irakobra [83]3 years ago
6 0

Answer:

4.90 M

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = ? M

V₁ = 125.0 mL

M₂ = 4.56 M

V₂ = 134.1 mL

Using the above formula , the molarity of acid , can be calculated as ,

M₁V₁ = M₂V₂  

Substituting the respective values ,  

M₁ *  125.0 mL = 4.56 M *  134.1 mL

M₁ = 4.90 M

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As a technician in a large pharmaceutical research firm, you need to produce 250. mL of a potassium dihydrogen phosphate buffer
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Answer:

We will need 147.772 mL of KH2PO4 to make this solution

Explanation:

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0.6918 = [HPO42-]/[H2PO4-]

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If we plug this in the equation 0.6918 = [HPO42-]/[H2PO4-]

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