What is the pH of a 1.0 L buffer made with 0.300 mol of HF (Ka = 6.8 × 10⁻⁴) and 0.200 mol of NaF to which 0.150 mol of HCl were added

Answers

Answer 1

Answer:

pH = 2.21

Explanation:

Hello there!

In this case, according to the reaction between NaF and HCl as the latter is added to the buffer:

[tex]NaF+HCl\rightarrow NaCl+HF[/tex]

It is possible for us to see how more HF is formed as HCl is added and therefore, the capacity of this HF/NaF-buffer is diminished as it turns acid. Therefore, it turns out feasible for us to calculate the consumed moles of NaF and the produced moles of HF due to the change in moles induced by HCl:

[tex]n_{HF}^{new}=0.300mol+0.150mol=0.450mol\\\\n_{NaF}^{new}=0.200mol-0.150mol=0.050mol[/tex]

Next, we calculate the resulting concentrations to further apply the Henderson-Hasselbach equation:

[tex][HF]=\frac{0.450mol}{1.0L} =0.450M[/tex]

[tex][NaF]=\frac{0.050mol}{1.0L} =0.050M[/tex]

Now, calculated the pKa of HF:

[tex]pKa=-log(6.8x10^{-4})=3.17[/tex]

We can proceed to the HH equation:

[tex]pH=pKa+log(\frac{[NaF]}{[HF]} )\\\\pH=3.17+log(\frac{0.05M}{0.45M} )\\\\pH=2.21[/tex]

Best regards!

Answer 2

pH is the measure of the hydrogen or hydronium ion concentration in an aqueous solution. The pH of the buffer made with HF and NaF is 2.21.

What is pH?

pH is the measure of the acidic or the basic content in a solution that can be given by the hydrogen or the hydroxide concentration.

The reaction can be shown as,

[tex]\rm NaF + HCl \rightarrow NaCl + HF[/tex]

New moles of HF is 0.300 + 0.150 = 0.450 moles and new moles of NaF is 0.200 - 0.150 = 0.050 moles.

The concentration is calculated by the Henderson-Hasselbach equation:

[tex]\rm [HF] = \dfrac{0.450 \;\rm mol}{1.0 \;\rm L} = 0.450 \;\rm M[/tex]

And,

[tex]\rm [NaF] = \dfrac{0.050\;\rm mol}{1.0 \;\rm L} = 0.050\;\rm M[/tex]

pKa of hydrogen fluoride is calculated as:

[tex]\rm pKa = - log (6.8 \times 10^{-4}) = 3.17[/tex]

The pH from the pKa can be calculated as:

[tex]\begin{aligned}\rm pH &=\rm pKa + log (\dfrac{[NaF]}{[HF]})\\\\&= 3.17 + \rm log (\dfrac{0.05}{0.45})\\\\&= 2.21\end{aligned}[/tex]

Therefore, the pH is 2.21.

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THIS IS A THREE PART QUESTION IF YOU CAN HELP IT WOULD BE REALLY APPRECIATED SO I DONT FAIL.

1. If I add 2.65L of water to 245 grams of sodium acetate, what is the molarity of the NaC2H3O2 solution?
A. 10.8 m
B. 1.13 m
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Answers

Answer:

1. B = 1.13M

2. A. 8.46%

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Explanation:

1. Molarity of a a solution = number of moles of solute/ volume of solution in L

Number of moles of solute = mass of solute/molar mass of solute

Molar mass of NaC₂H₃O₂ = 82 g/mol, mass of NaC₂H₃O₂ = 245 g

Number of moles of NaC₂H₃O₂ = 245 g / 82 g/mol = 2.988 moles

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2. Percentage by mass of a substance = mass of substance /mass of solution × 100%

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Density of water = 1 Kg/L = 1000 g/L; volume of waterb= 2.65 L

Mass of water = 1000 g/L × 2.65 L = 2650 g

Mass of solution = mass of water + mass of solute = 245 + 2650 =2895 g

Percentage by mass of NaC₂H₃O₂ = 245/2895 × 100% = 8.46%

3. Mole fraction of NaC₂H₃O₂ = moles of NaC₂H₃O₂/moles of solution

Moles of water = mass /molar mass

Mass of water = 2650 g; molar mass of water = 18 g/mol

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zinc + hydrogen arsenate—>

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what did you get?

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Answers

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Given :-

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sry i its too long

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