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andriy [413]
3 years ago
10

Calculate the pH at the equivalence point when 22.0 mL of 0.200 M hydroxylamine, HONH2, is titrated with 0.15 M HCl. (Kb for HON

H2
Chemistry
1 answer:
solniwko [45]3 years ago
4 0

Answer:

pH = 3.513

Explanation:

Hello there!

In this case, since this titration is carried out via the following neutralization reaction:

HONH_2+HCl\rightarrow HONH_3^+Cl^-

We can see the 1:1 mole ratio of the acid to the base and also to the resulting acidic salt as it comes from the strong HCl and the weak hydroxylamine. Thus, we first compute the required volume of HCl as shown below:

V_{HCl}=\frac{22.0mL*0.200M}{0.15M}=29.3mL

Now, we can see that the moles of acid, base and acidic salt are all:

0.0220L*0.200mol/L=0.0044mol

And therefore the concentration of the salt at the equivalence point is:

[HONH_3^+Cl^-]=\frac{0.0044mol}{0.022L+0.0293L} =0.0858M

Next, for the calculation of the pH, we need to write the ionization of the weak part of the salt as it is able to form some hydroxylamine as it is the weak base:

HONH_3^++H_2O\rightleftharpoons H_3O^++HONH_2

Whereas the equilibrium expression is:

Ka=\frac{[H_3O^+][HONH_2]}{[HONH_3^+]}

Whereas Ka is computed by considering Kw and Kb of hydroxylamine:

Ka=\frac{Kw}{Kb}=\frac{1x10^{-14}}{9.10x10^{-9}}  \\\\Ka=1.10x10^{-6}

So we can write:

1.10x10^{-6}=\frac{x^2}{0.0858-x}

And neglect the x on bottom to obtain:

1.10x10^{-6}=\frac{x^2}{0.0858}\\\\x=\sqrt{1.10x10^{-6}*0.0858}=3.07x10^{-4}M

And since x=[H3O+] we obtain the following pH:

pH=-log(3.07x10^{-4})\\\\pH=3.513

Regards!

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Answer:

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General Formulas and Concepts:

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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<u>Chemistry</u>

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Explanation:

<u>Step 1: Define</u>

0.200 mol H₂S

<u>Step 2: Identify Conversions</u>

Molar Mass of H - 1.01 g/mol

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Molar Mass of H₂S - 2(1.01) + 32.07 = 34.09 g/mol

<u>Step 3: Convert</u>

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Percentage yield: 54.5%

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Theoretical yield of para-nitroaniline is the mass produced assuming a yield of 100%. That is:

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Assuming a yield of 100% are 0.020 moles of paranitroaniline. Theoretical yield is:

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<h3>Theoretical yield: 2.75g of paranitroaniline</h3><h3 />

Percentage yield is:

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Actual yield was 1.5g and percentage yield will be:

Percentage yield: (1.5g / 2.75g) × 100

<h3>Percentage yield: 54.5%</h3>

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