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A student dissolved 1.805g of a monoacidic weak base in 55mL of water. Calculate the equilibrium pH for the weak monoacidic base (B) solution. Show all your work.pKb for the weak base = 4.82.

Molar mass of the weak base = 82.0343g/mole.

Note: pKa = -logKa

pKb = -logKb

pH + pOH = 14

[H+ ] [OH- ] = 10^-14

Answers

Answer 1

Answer:

Answer:

11.39

Explanation:

Given that:

$pK_(b)=4.82$

$K_(b)=10^(-4.82)=1.5136* 10^(-5)$

Given that:

Mass = 1.805 g

Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

$moles = (Mass\ taken)/(Molar\ mass)$

Thus,

$Moles= (1.805\ g)/(82.0343\ g/mol)$

$Moles= 0.022\ moles$

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

$Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)$

$Molarity=(0.022)/(0.055)$

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

B + H₂O ⇄ BH⁺ + OH⁻

At t=0 0.4 - -

At t =equilibrium (0.4-x) x x

The expression for dissociation constant is:

$K_(b)=\frac {\left [ BH^(+) \right ]\left [ {OH}^- \right ]}{[B]}$

$1.5136* 10^(-5)=\frac {x^2}{0.4-x}$

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³ M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

pH = 14 - pOH = 14 - 2.61 = 11.39

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Answers

Answer:

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Cyclohexanecarboxylic acid, C6H11COOH (pKa 4.90), is only slightly soluble in water, but its sodium salt, C6H11COO-Na , is quite soluble in water. Describe the solubility of cyclohexanecarboxylic acid in solutions of sodium hydroxide, sodium bicarbonate (NaHCO3), and sodium carbonate (Na2CO3). The pKa values for the conjugate acids of sodium hydroxide, sodium bicarbonate (NaHCO3), and sodium carbonate (Na2CO3) are 15.7, 6.36, and 10.33, respectively.

Answers

Answer:

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Explanation:

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Answers

Answer:

0.113 M

Explanation:

Since B and D are on opposite sides of the reaction, the concentration of D increases when the concentration of B decreases. The amount by which D increases is determined by the coefficients of B and D in the balanced chemical equation:

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A structural model of retinol is shown below. How many carbon atoms are inretinol?
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Answers

Answer:

The answer is 20

Explanation:

Assuming 100% dissociation, which of the following compounds is listed incorrectly with its van't Hoff factor i? Al2(SO4)3, i = 4 NH4NO3, i = 2 Mg(NO3)2, i = 3 Na2SO4, i = 3 Sucrose, i = 1

Answers

Answer:

- Aluminium sulfate Al2(SO4)3 dissociates in two aluminium ions and three sulfate ions, therefore, van't Hoff factor is 5 (incorrect).

Explanation:

Hello,

In this case, since the van't Hoff factor is related with the species that result from the ionization of a chemical compound, we can see that that

- Aluminium sulfate Al2(SO4)3 dissociates in two aluminium ions and three sulfate ions, therefore, van't Hoff factor is 5 (incorrect).

- Ammonium nitrate NH4NO3 dissociates in one ammonium ions and one nitrate ion, therefore, van't Hoff factor is 2 (correct).

- Sodium sulfate Na2SO4 dissociates in two sodium ions and one sulfate, therefore, van't Hoff factor is 3 (correct).

- Sucrose is not ionized, therefore, van't Hoff factor is 1 (correct).

Best regards.

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