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Papessa [141]
3 years ago
13

The titration of 78.5 mL of an unknown concentration H3PO4 solution requires 134 mL of 0.224 M KOH solution. What is the concent

ration of the H3PO4 solution
Chemistry
1 answer:
Katyanochek1 [597]3 years ago
3 0

Answer:

0.127 M.

Explanation:

The balanced equation for the reaction is given below:

H3PO4 + 3KOH —› K3PO4 + 3H2O

From the balanced equation above, we obtained the following data:

Mole ratio of acid, H3PO4 (nA) = 1

Mole ratio of base, KOH (nB) = 3

Data obtained from the question include:

Volume of acid, H3PO4 (Va) = 78.5 mL

Molarity of acid, H3PO4 (Ma) =...?

Volume of base, KOH (Vb) = 134 mL

Molarity of base, KOH (Mb) = 0.224 M

The concentration of the acid, H3PO4 can be obtained as follow:

MaVa / MbVb = nA/nB

Ma x 78.5 / 0.224 x 134 = 1/3

Cross multiply

Ma x 78.5 x 3 = 0.224 x 134 x 1

Divide both side by 78.5 x 3

Ma = (0.224 x 134 x 1) /(78.5 x 3)

Ma = 0.127 M

Therefore, the concentration of the acid, H3PO4 is 0.127 M.

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2NO(g)+5H_2(g)\rightarrow 2NH_3(g)+2H_2O(g)

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As, 2 mole of NO react to give 2 mole of NH_3

So, 1.53 mole of NO react to give 1.53 mole of NH_3

Now we have to calculate the mass of NH_3

\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3

\text{ Mass of }NH_3=(1.53moles)\times (17g/mole)=26.0g

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