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marin [14]
3 years ago
13

If 6.89 g of CuNO3 is dissolved in water to make a 0.460 M solution, what is the volume of the solution?

Chemistry
1 answer:
Readme [11.4K]3 years ago
6 0
3.67  grams is the volume of the solution
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An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni
Svetllana [295]

<u>Answer:</u> The pH of acid solution is 4.58

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}    .....(1)

  • <u>For KOH:</u>

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol

  • <u>For propanoic acid:</u>

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol

The chemical reaction for propanoic acid and KOH follows the equation:

                 C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O

<u>Initial:</u>          0.1372         0.04514  

<u>Final:</u>           0.09206          -                0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L     (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})

pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})

We are given:  

pK_a = negative logarithm of acid dissociation constant of propanoic acid = 4.89

[C_2H_5COOK]=\frac{0.04514}{0.26594}

[C_2H_5COOH]=\frac{0.09206}{0.26594}

pH = ?  

Putting values in above equation, we get:

pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58

Hence, the pH of acid solution is 4.58

7 0
3 years ago
Concentrated HCl is 37% m/m. The density of the solution is 1.19 g/mL. Assume you have 100.0 grams of SOLUTION.
sdas [7]

Answer:

M HCl sln = 12.0785 M

Explanation:

  • molarity (M) [=] mol/L
  • %mm = ((mass compound)/(mass sln))*100

∴ mass sln = 100.0 g

∴ δ sln = 1.19 g/mL

∴ % m/m = 37 %

⇒ 37 % =((mass HCl/mass sln))*100

⇒ 0.37 = mass HCl / 100.0 g

⇒ 37 g = mass HCl

∴ molar mass HCl = 36.46 g/mol

⇒ mol HCl = (37 g)*(mol/36.46 g) = 1.015 mol

⇒ volume sln = (100 g sln)*(mL/1.19 g) =  84.034 mL = 0.084034 L

⇒ M HClsln = 1.015 mol/0.084034 L

⇒ M HCl sln = 12.0785 M

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Answer:

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