15) What is the hydronium ion concentration [H3O + ] of a 0.125 M hydrocyanic acid solution with Ka = 4.9 × 10-10? The equation for the dissociation of hydrocyanic acid is: HCN(aq) + H2O(l) ⇌ H3O+(aq) + CN- (aq)

Answer:

c. CH3CH2CH(CH3)CHO

Explanation:

Hello there!

In this case, according to the process for the one-step oxidation of a primary alcohol with a moderately strong oxidizing agent like pyridinium chlorochromate (PCC), whereby an aldehyde is produced, we infer that the corresponding product will be 2-methylbutanal, which matches with the choice c. CH3CH2CH(CH3)CHO according to the following reaction:

Regards!

Answer:

i) pH = 0.6990

ii) pH = 2.389

Explanation:

i) Before adding aqueous NaOH, there are 25.00 mL of 0.2000 M HCl. HCl reacts with the water in the aqueous solution as follows:

HCl + H₂O ⇒ H₃O⁺ + Cl⁻

The HCl and H₃O⁺ are related to each other through a 1:1 molar ratio, so the concentration of H₃O⁺ is equal to the HCl concentration.

The pH is related to the hydronium ion concentration as follows:

pH = -log([H₃O⁺]) = -log(0.2000) = 0.699

ii) Addition of NaOH causes the following reaction:

H₃O⁺ + NaOH ⇒ 2H₂O + Na⁺

The H₃O⁺ and NaOH react in a 1:1 molar ratio. The amount of NaOH added is calculated:

n = CV = (0.2000 mol/L)(24.00 mL) = 4.800 mmol NaOH

Thus, 4.800 mmol of H₃O⁺ were neutralized.

The initial amount of H₃O⁺ present was:

n = CV = (0.2000 mol/L)(25.00 mL) = 5.000 mmol H₃O⁺

The amount of H₃O⁺ that remains after addition of NaOH is:

(5.000 mmol) - (4.800 mmol) = 0.2000 mmol

The concentration of H₃O⁺ is the amount of H₃O⁺ divided by the total volume. The total volume is (25.00 mL) + (24.00 mL ) = 49.00 mL

C = n/V = (0.2000 mmol) / (49.00 mL) = 0.004082 M

The pH is finally calculated:

pH = -log([H₃O⁺]) = -log(0.004082) = 2.389

Solution : Water and Methanol are easily miscible in any amount. so they are not preferred for the liquid-liquid extraction process.

Liquid-Liquid Extraction is also called as solvent extraction. It is the method of seperation of compound based on their relative solubilities in two different immiscible liquids. Generally we use water (polar) and an organic solvent (non-polar).

It is important that the two solvents should not be mix because it is easy to seperate them.

Water and Methanol are easily miscible in any amount. we can not seperate them easily. So that is why we can not use water and methanol as a solvent in liquid-liquid extraction process.

Answer:

they could be joined due to hydrogen bridge-type intermolecular interactions so no phase splitting will be carried out.

Explanation:

Hello,

Liquid-liquid extraction is a widely used separation operation that is suitable when relative volatilities are so close, so an extra substance is used to modify the equilibrium causing a phase splitting (two liquid immiscible phases) which could be leveraged to mechanically separate the two phases. The basic idea lies on the fact that the extra substance must be largely immiscible with the original solvent, to the solute is selectively separated, nonetheless, in this case, water and methanol are largely soluble to each other since they could be joined due to hydrogen bridge-type intermolecular interactions so no phase splitting will be carried out.

Best regards.

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

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:

brainly.com/question/29760073

#SPJ3

Answer:

a. Acid B

b. Acid B

c. lower

Hope this helps you