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Blababa [14]
3 years ago
7

A 25.225 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 75.815 g of wa

ter. A 14.842 g aliquot of this solution is then titrated with 0.1068 M HCl . It required 29.99 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste
Chemistry
1 answer:
Rudik [331]3 years ago
6 0

Answer:

1.43 (w/w %)

Explanation:

HCl reacts with NH3 as follows:

HCl + NH3 → NH4+ + Cl-

<em>1 mole of HCl reacts per mole of ammonia.</em>

Mass of NH3 is obtained as follows:

<em>Moles HCl:</em>

0.02999L * (0.1068mol / L) = 3.203x10-3 moles HCl = <em>Moles NH3</em>

<em>Mass NH3 in the aliquot:</em>

3.203x10-3 moles NH3 * (17.031g / mol) = 0.0545g.

Mass of sample + water = 22.225g + 75.815g = 98.04g

Dilution factor: 98.04g / 14.842g = 6.6056

That means mass of NH3 in the sample is:

0.0545g * 6.6056 = 0.36g NH3

Weight percent is:

0.36g NH3 / 25.225g * 100

<h3>1.43 (w/w %)</h3>
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<u>Explanation:</u>

To calculate the molarity of NiC_2O_4, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of NiC_2O_4 = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M

For the given chemical equations:

NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}

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The expression for equilibrium constant of above equation is:

K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}

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We are given:

[[Ni(NH_3)_6]^{2+}]=0.016M

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