Chemistry, 03.08.2020 14:01 taykola

How many grams of dipotassium succinate trihydrate (K2C4H4O4·3H2O, MW = 248.32 g/mol) must be added to 660.0 mL of a 0.0577 M succinic acid solution to produce a pH of 5.869 ? Succinic acid has pKa values of 4.207 (pKa1) and 5.636 (pKa2).

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Answer from: joannachavez12345

the amount of grams of dipotassium succinate trihydrate K₂C₄H₄O₄·3H₂O = 41.798g

Explanation:

Given that:

The volume of K_2C_4H_4O_4.3H_2O = 660.mL

The molarity of succinic acid = 0.0577 M

The pH of the solution = 5.869

The pKa₁ = 4.207

The pKa₂ = 5.636

The pKa₂ is required to be used for the determination of given that succinic acid is dissociated twice.

Using Henderson-HasselBalch Equation,

$PH =pKa + log\dfrac{[Salt]}{[Acid]}$

we know that :

The volume of K_2C_4H_4O_4 = 660.mL

The molarity of succinic acid = 0.0577 M

number of moles = Molarity $\times$ Volume (liters)

number of moles = 0.0577 $\times$ 0.660 L

number of moles = 0.038082 mol

The balanced chemical reaction for this equation can be expressed as follows:

$K_2C_4H_4O_4+C_4H_4O_4^{2-} \longleftrightarrow 2KC_4H_4O_4^-$

Here, the number of moles of $K_2C_4H_4O_4= C_4H_4O_4^{2-}$

number of moles of 2KC_4H_4O_4^- = 2 × 0.038082 mol

number of moles of 2KC_4H_4O_4^- = 0.076164 mol

From the Henderson-HasselBalch Equation,

$pH =pKa + log\dfrac{[Salt]}{[Acid]}$

$pH =pKa_2 + log\dfrac{[K_2C_4H_4O_4]}{[KC_4H_4O_4^-]}$

$pH =pKa_2 + log\dfrac{n_{K_2C_4H_4O_4}}{n_{KC_4H_4O_4^-}}$

$5.869 =5.636+ log \dfrac{n_{K_2C_4H_4O_4}}{0.076164 }$

$0.233= log \dfrac{n_{K_2C_4H_4O_4}}{0.076164 }$

$10^{0.233}= \dfrac{n_{K_2C_4H_4O_4}}{0.076164 }$

$1.710015315= \dfrac{n_{K_2C_4H_4O_4}}{0.076164 }$

$1.710015315 \times 0.076164= n_{K_2C_4H_4O_4$

$n_{K_2C_4H_4O_4}= 0.1302416$

SInce; the number of moles of $K_2C_4H_4O_4= C_4H_4O_4^{2-}$

$n_{C_4H_4O_4^{2-}} =$ ( 0.1302416 + 0.038082) mol

$n_{C_4H_4O_4^{2-}} =$ 0.1683236 mol

The grams of dipotassium succinate trihydrate = numbers of moles of dipotassium succinate trihydrate × Molar mass

The grams of dipotassium succinate trihydrate = 0.1683236 mol × 248.32 g/mol

The grams of dipotassium succinate trihydrate = 41.798g

Thus , the amount of grams of dipotassium succinate trihydrate K₂C₄H₄O₄·3H₂O = 41.798g

Answer from: Quest

the grams of barium that are in 2.3 x 10^23 atoms of barium is 52.4 g (answer c)

calculation

mass = moles of ba x molar mass of ba

molar mass of ba from periodic table= 137.3 g/mol

use the avogadro's law constant to calculate the moles of barium

that is 1 mole = 6.02 x 10^23 atoms

? moles = 2.3 x 10^23 atoms

by cross multiplication

=(1 mole x 2.3x 10^23 atoms) / 6.02 x10^23 atoms)= 0.382 moles

mass is therefore = 137.3 g/mol x 0.382 mol = 52.4 ( answer c)

explanation:

Answer from: Quest

Hcn is the correct answer leave a good rate and .

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How many grams of dipotassium succinate trihydrate (K2C4H4O4·3H2O, MW = 248.32 g/mol) must be added...

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