A chemist titrates 60.0 mL of a 0.1935 M benzoic acid (HC (H5CO2) solution with 0.2088 M KOH solution at 25 °C. Calculate the pH at equivalence. The pKg of benzoic acid is 4.20.

Answers

Answer 1

Answer:

Answer:

pH at the equivalence point is 8.6

Explanation:

A titulation between a weak acid and a strong base, gives a basic pH at the equivalence point. In the equivalence point, we need to know the volume of base we added, so:

mmoles acid = mmoles of base

60 mL . 0.1935M = 0.2088 M . volume

(60 mL . 0.1935M) /0.2088 M = 55.6 mL of KOH

The neutralization is:

HBz + KOH ⇄ KBz + H₂O

In the equilibrum:

HBz + OH⁻ ⇄ Bz⁻ + H₂O

mmoles of acid are: 11.61 and mmoles of base are: 11.61

So in the equilibrium we have, 11.61 mmoles of benzoate.

[Bz⁻] = 11.61 mmoles / (volume acid + volume base)

[Bz⁻] = 11.61 mmoles / 60 mL + 55.6 mL = 0.100 M

The conjugate strong base reacts:

Bz⁻ + H₂O ⇄ HBz + OH⁻ Kb

0.1 - x x x

(We don't have pKb, but we can calculate it from pKa)

14 - 4.2 = 9.80 → pKb → 10⁻⁹'⁸ = 1.58×10⁻¹⁰ → Kb

Kb = [HBz] . [OH⁻] / [Bz⁻]

Kb = x² / (0.1 - x)

As Kb is so small, we can avoid the quadratic equation

Kb = x² / 0.1 → Kb . 0.1 = x²

√ 1.58×10⁻¹¹ = [OH⁻] = 3.98 ×10⁻⁶ M

From this value, we calculate pOH and afterwards, pH (14 - pOH)

- log [OH⁻] = pOH → - log 3.98 ×10⁻⁶ = 5.4

pH = 8.6

Answer 2

Answer:

Final answer:

To calculate the pH at equivalence in a titration, we need to consider the concentration of the excess strong base in the solution. First, we calculate the moles of the acid and the base, then we find the moles of the excess base. Using this information, we can find the concentration of the excess base and subsequently calculate pOH. Finally, we can convert pOH to pH using the pH + pOH = 14 relationship.

Explanation:

pH at the equivalence point in a titration can be determined by considering the concentration of the excess strong base present in the reaction mixture. In this case, the excess strong base is KOH. We can calculate [OH-] using the stoichiometry of the reaction and the given concentrations. Then, we can find the pOH using the formula -log[OH-]. Finally, we can convert pOH to pH using the pH + pOH = 14 relationship.

Given:

Volume of benzoic acid solution (HC (H5CO2)): 60.0 mL
Concentration of benzoic acid solution: 0.1935 M
Concentration of KOH solution: 0.2088 M

Step 1: Determine the amount of benzoic acid (HC (H5CO2)) in moles:

moles of HC (H5CO2) = volume (L) × concentration (M) = 0.0600 L × 0.1935 M = 0.01161 mol

Step 2: Determine the amount of KOH in moles:

moles of KOH = volume (L) × concentration (M) = 0.0600 L × 0.2088 M = 0.01253 mol

Step 3: Determine the amount of excess KOH in moles:

moles of excess KOH = moles of KOH - moles required for neutralizing HC (H5CO2) = 0.01253 mol - 0.01161 mol = 9.2 × 10-4 mol

Step 4: Determine the concentration of excess KOH:

concentration of excess KOH = moles of excess KOH / volume (L) = 9.2 × 10-4 mol / 0.0600 L = 0.0153 M

Step 5: Determine the pOH of the solution:

pOH = -log[OH-] = -log(0.0153) ≈ 1.82

Step 6: Determine the pH of the solution:

pH = 14 - pOH = 14 - 1.82 ≈ 12.18

Learn more about pH at equivalence point here:

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To what extend must a soln conc 40mg 4g NO3 pa cm3 be dilute to yield one of conc 16mg 4g NO3 pa cm3

Answers

Each ml should be dilute to 2.5 ml

To each ml of solution 1.5 ml of water should be added

When 282 gr of glycine are dissolved in 950 gr of a certain mystery liquid, the freezing point of the solution is 8.2 C less than the freezing point of pure. Calculate the mass of iron(III) chloride that must be dissolved in the same mass of to produce the same depression in freezing point. The van't Hoff factor for iron(III) chloride in X. Be sure your answer has a unit symbol, if necessary, and round your answer to significant digits.

Answers

Answer: $1.4* 10^2g$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i* K_f* m$

$\Delta T_f=T_f^0-T_f=8.2^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte like glycine)

$K_f$ = freezing point constant = ?

m= molality

$\Delta T_f=i* K_f* \frac{\text{mass of solute}}{\text{molar mass of solute}* \text{weight of solvent in kg}}$

Weight of solvent = 950 g = 0.95 kg

Molar mass of glycine = 75.07 g/mol

Mass of glycine added = 282 g

$8.2=1* K_f* (282g)/(75.07 g/mol* 0.95kg)$

$K_f=2.2^0C/m$

Thus freezing point constant is $2.2^0C/m$

2) $\Delta T_f=i* K_f* m$

$\Delta T_f=T_f^0-T_f=8.2^0C$ = Depression in freezing point

i= vant hoff factor = 4 (for $FeCl_3$ )

$K_f$ = freezing point constant = $2.2C/m$

m= molality

$\Delta T_f=i* K_f* \frac{\text{mass of solute}}{\text{molar mass of solute}* \text{weight of solvent in kg}}$

Weight of solvent = 950 g = 0.95 kg

Molar mass of $FeCl_3$ = 162.2 g/mol

Mass of $FeCl_3$ added = ?

$8.2=4* 2.2* (xg)/(162.2 g/mol* 0.95kg)$

$x=1.4* 10^2g$

Thus mass of iron(III) chloride that must be dissolved in the same mass of to produce the same depression in freezing point is $1.4* 10^2g$

Answer:

$mass_(FeCl_3)=1.5x10^2gFeCl_3$

Explanation:

Hello,

In this case, by using the given data for glycine, one computes the freezing point constant of the mystery liquid as shown below, considering the molality of the glycine and its van't Hoff factor equal to the unity:

$\Delta T=i*Kf*m_(Glyc)\n\nKf=(\Delta T)/(i*m_(glyc)) =(8.2^oC)/(1*(282gGlyc)/(950gX)*(1molGlyc)/(75.07gGlyc)*(1000gX)/(1kgX) ) \n\nKf=2.1^oC/m$

Now, as we are looking for the mass of iron(III) chloride at the same conditions of the aforesaid case, at first, one solves for the molarity of such compound considering that its theoretical van't Hoff factor is 4 as follows:

$m_(FeCl_3)=(\Delta T)/(i*Kf) =(8.2^oC)/(4*2.1^oC/m_(FeCl_3)) =0.98m$

Now, one obtains the requested mass via:

$mass_(FeCl_3)=0.98(molFeCl_3)/(kgX)*0.95kgX*(162.35gFeCl_3)/(1molFeCl_3) \n\nmass_(FeCl_3)=151.1gFeCl_3\nmass_(FeCl_3)=1.5x10^2gFeCl_3$

Best regards.

A substance with a high boiling point will also likely have A) a low melting point B) a low vapor pressure C) weak intermolecular interactions D) low surface tension E) low viscosity

Answers

Answer: Option (B) is the correct answer.

Explanation:

Boiling point is defined as the temperature at which vapor pressure of a liquid becomes equal to the atmospheric pressure.

Surface tension is defined as the attractive forces experienced by the surface molecules of a liquid by the molecules present beneath the surface layer of the liquid. More stronger are the intermolecular forces present within the molecules of a liquid more will be its surface tension and lower will be its vapor pressure.

Hence, higher will be the boiling point of the liquid.

And, viscosity is defined as the ability of a liquid to resist its flow. When a substance has high viscosity then it is known as a viscous substance.

Thus, we can conclude that a substance with a high boiling point will also likely have a low vapor pressure.

What is the percent yield if 23.1 grams of FeCl3 (162.2 g/mol) is created from 10.61 grams of iron (55.85 g/mol) in the following reaction Report your answer with three significantfigures
2 Fe + 3Cl2 + 2 FeCl3

Answers

The percentage yield of the reaction if 23.1 grams of FeCl₃ is created by 10.61g of Fe is 74.9 %.

What is percent yield?

Percent yield of a reaction is the ratio of actual yield to the theoretical yield multiplied by 100.

As per the given balanced reaction, one mole or 55.85 g of Fe is needed to produce 1 mole or 162.2 g of FeCl₃.

Then theoretically, the mass of FeCl₃ which can be produced by 10.61 g of Fe is calculated as follows:

= (10.61 × 162.2) / 55.85

=30.81 g

This is the theoretical yield.

The actual yield is given 23.1. Now the percentage yield is calculated as follows:

Percentage yield = (Actual yield/ heretical yield)×100

= (23.1/30.81)×100

= 74.9 %.

Hence, the percentage yield of the reaction is 74.9 %.

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Similar question just through and use it to answer your. Thanks. Hope it helps.

With 21 g of Zinc, and 7 g of CuCl2, how much ZnCl2 is made in grams?

Answers

Answer: 7.07 grams

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}*{\text{Molar Mass}}$

$\text{Moles of} zinc=(21g)/(65g/mol)=0.32moles$

$\text{Moles of} CuCl_2=(7g)/(134g/mol)=0.052moles$

$Zn+CuCl_2\rightarrow Cu+ZnCl_2$

According to stoichiometry :

1 mole of $CuCl_2$ require 1 mole of $Zn$

Thus 0.052 moles of $CuCl_2$ will require= $(1)/(1)* 0.052=0.052moles$ of $Zn$

Thus $CuCl_2$ is the limiting reagent as it limits the formation of product and $Zn$ is the excess reagent.

As 1 mole of $CuCl_2$ give = 1 mole of $ZnCl_2$

Thus 0.052 moles of $CuCl_2$ give = $(1)/(1)* 0.052=0.052moles$ of $ZnCl_2$

Mass of $ZnCl_2=moles* {\text {Molar mass}}=0.052moles* 136g/mol=7.07g$

Thus 7.07 g of $ZnCl_2$ will be produced from the given masses of both reactants.

Consider the generic chemical equation below. X + Ymc021-1.jpg W + Z Reactant X contains 199.3 J of chemical energy. Reactant Y contains 272.3 J of chemical energy. Product W contains 41.9 J of chemical energy. If the reaction loses 111.6 J of chemical energy as it proceeds, how much chemical energy must product Z contain?