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nikdorinn [45]
3 years ago
7

At 22 °C, an excess amount of a generic metal hydroxide, M(OH)2, is mixed with pure water. The resulting equilibrium solution ha

s a pH of 10.22. What is the Ksp of the salt at 22 °C?
Chemistry
1 answer:
brilliants [131]3 years ago
7 0

Answer:

2.29x10⁻¹² is Ksp of the salt

Explanation:

The Ksp of the metal hydroxide is:

M(OH)₂(s) ⇄ M²⁺ + 2OH⁻

Ksp = [M²⁺] [OH⁻]²

As you can see in the reaction, 2 moles of OH⁻ are produced per mole of M²⁺. It is possible to find [OH⁻] with pH, thus:

pOH = 14- pH

pOH = 14 - 10.22

pOH = 3.78

pOH = -log[OH⁻]

<em>1.66x10⁻⁴ = [OH⁻]</em>

And [M²⁺] is the half of [OH⁻], <em>[M²⁺] = 8.30x10⁻⁵</em>

<em />

Replacing in Ksp formula:

Ksp = [8.30x10⁻⁵] [1.66x10⁻⁴]²

Ksp = 2.29x10⁻¹² is Ksp of the salt

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A buffer contains 0.19 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. You may want to
Snezhnost [94]

Explanation:

It is known that pK_{a} of propionic acid = 4.87

And, initial concentration of  propionic acid = \frac{0.19}{1.20}

                                                                       = 0.158 M

Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

                                                             = 0.216 M

Now, in the given situation only propionic acid and sodium propionate are present .

Hence,      pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.216}{0.158}

                        = 4.87 + log (1.36)

                        = 5.00

  • Therefore, when 0.02 mol NaOH is added  then,

     Moles of propionic acid = 0.19 - 0.02

                                              = 0.17 mol

Hence, concentration of propionic acid = \frac{0.17}{1.20 L}

                                                                 = 0.14 M

and,      moles of sodium propionic acid = (0.26 + 0.02) mol

                                                                  = 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.28 mol}{1.20 L}

                           = 0.23 M

Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.23}{0.14}

                        = 4.87 + log (1.64)

                        = 5.08

Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.

  • Therefore, when 0.02 mol HI is added  then,

     Moles of propionic acid = 0.19 + 0.02

                                              = 0.21 mol

Hence, concentration of propionic acid = \frac{0.21}{1.20 L}

                                                                 = 0.175 M

and,      moles of sodium propionic acid = (0.26 - 0.02) mol

                                                                  = 0.24 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.24 mol}{1.20 L}

                           = 0.2 M

Therefore, calculate pH upon addition of 0.02 mol of HI as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.2}{0.175}

                        = 4.87 + log (0.114)

                        = 4.98

Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.

7 0
3 years ago
1.) Calculation: If 9.02 x 1024 particles of vinegar (HC2H3O2)HC2H3O2) are added to 16.5 moles of eggshell (CaCO3) and 6.35 mole
blsea [12.9K]

The theoretical yield of acetate is 2607 g. The actual yield of acetate is 1066.8 g. The percentage yield of acetate is 41%.

If 1 mole of vinegar contains 6.02  x 10^23 particles

x moles of vinegar contains 9.02 x 10^24 particles

x = 1 mole x 9.02 x 10^24 /6.02 x 10^23

x = 15 moles of vinegar

The reaction is as follows;

2HC2H3O2 + CaCO3 -----> Ca(C2H3O2)2 + H2O + CO2

Since 2 moles of vinegar reacts with 1 mole of carbonate

x moles of vinegar reacts with 16.5 moles of carbonate

x =  2 moles x 16.5 moles/ 1 mole

x = 33 moles of vinegar

We can see that the vinegar is the reactant in excess hence the carbonate is the limiting reactant.

Theoretical yield = 16.5 moles x 158 g/mol = 2607 g

Actual yield = 6.35 moles  x 158 g/mol = 1066.8 g

Percent yield = 1066.8 g/2607 g × 100/1

= 41%

Learn more: brainly.com/question/13440572?

7 0
3 years ago
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