4. Calculate the [𝐻^+ ] 𝑎𝑛𝑑 [𝑁𝑂_3^( −) ] of a 0.175M solution of nitric acid, 𝑁𝐻𝑂_3^( −).

Question

4. Calculate the [𝐻^+ ] 𝑎𝑛𝑑 [𝑁𝑂_3^( −) ] of a 0.175M solution of nitric acid, 𝑁𝐻𝑂_3^( −).

5. Calculate the hydroxide ion concentration in a 0.175 solution of HCl.
6. Calculate the hydrogen ion concentration in 0.01M solution of Ca(𝑂𝐻)_2
7. The ionisation constant of water is 𝐾_𝑤=1.00×10^(−13.60) at body temperature, 37^0 𝐶. 𝑊ℎ𝑎𝑡 𝑎𝑟𝑒 𝑡ℎ𝑒 𝐻_3 𝑂^+ 𝑎𝑛〖 𝑂𝐻〗^− concentrations at that temperature?
8. Calculate the pH of the following solutions at 25^0 𝐶:(𝑎)0.0028𝑀 𝐻𝐶𝑙, (𝑏)0.014 𝐻𝑁𝑂_3, (𝑐)0.00052𝑀 𝐻𝐵𝑟,
0.092𝑀 𝐻𝐼
9. Calculate the pH of a 0.025M solution of ammonia, 𝐾_𝑏=1.75× 10^(−5)

Answer 1

4. To calculate the [H+] and [NO3-] of a 0.175M solution of nitric acid, we need to consider that nitric acid is a strong acid and fully dissociates in water. Therefore, for every 1 mol of nitric acid, we will get 1 mol of H+ and 1 mol of NO3-.

Therefore, the [H+] and [NO3-] in the 0.175M solution of nitric acid will also be 0.175M.

[H+] = 0.175 M
[NO3-] = 0.175 M

5. Since HCl is also a strong acid, it fully dissociates in water. Therefore, the concentration of hydroxide ions (OH-) in a solution of HCl will be negligible.

6. According to the neutralization reaction between Ca(OH)2 and water, 1 mol of Ca(OH)2 produces 2 moles of OH-. Therefore, the concentration of OH- in a 0.01M solution of Ca(OH)2 will be double the concentration of Ca(OH)2.

[OH-] = 2 * 0.01 M = 0.02 M

7. The ionization constant of water, Kw, at 37°C is 1.00×10^-13.60. At equilibrium, the concentration of H3O+ (hydroxonium ions) will be equal to the concentration of OH-. Therefore, we can take the square root of Kw to calculate the concentrations of H3O+ and OH-.

[H3O+] = [OH-] = sqrt(Kw)
= sqrt(1.00×10^-13.60)

8.
(a) The pH of a solution can be calculated using the equation pH = -log[H+]. For 0.0028M HCl, the concentration of H+ is 0.0028M.

pH = -log(0.0028)

(b) For 0.014M HNO3, the concentration of H+ is 0.014M.

pH = -log(0.014)

(c) For 0.00052M HBr, the concentration of H+ is 0.00052M.

pH = -log(0.00052)

(d) For 0.092M HI, the concentration of H+ is 0.092M.

pH = -log(0.092)

9. The ionization constant of ammonia, Kb, is given as 1.75×10^-5. Since ammonia is a weak base, it does not completely dissociate in water. To calculate the pH of a 0.025M solution of ammonia, we need to consider the equilibrium of the reaction between ammonia and water: NH3 + H2O ⇌ NH4+ + OH-.

[H+] can be calculated using the Kb value:

Kb = [NH4+][OH-] / [NH3]

[H+] = sqrt(Kw * Kb * [NH3])

where [NH3] is the concentration of ammonia, which is 0.025M.

Then, the pH can be calculated as pH = -log[H+].

Answer 2

9. Calculate the pH of a 0.025M solution of ammonia, 𝐾_𝑏=1.75× 10^(−5)

Answer 3

To calculate the pH of a 0.025M solution of ammonia (NH3) with a Kb value of 1.75×10^(-5), we need to consider the equilibrium between ammonia and water:

NH3 + H2O ⇌ NH4+ + OH-

Since ammonia is a weak base, it will partially dissociate in water. Let x be the amount (in moles) of NH3 that dissociates.

The initial concentration of NH3 is 0.025M. After dissociation, the concentration of NH3 will be (0.025 - x)M, and the concentration of NH4+ and OH- will both be xM.

Since the equilibrium constant (Kb) is given by [NH4+][OH-]/[NH3], we can set up the equation:

Kb = [NH4+][OH-] / [NH3]
1.75×10^(-5) = x * x / (0.025 - x)

Now, we can solve this quadratic equation to find the value of x, and subsequently, the concentrations of NH4+ and OH-. Using the quadratic formula, we get:

x = [-(0.025) ± sqrt((0.025)^2 + 4 * 1.75×10^(-5) * 0.025)] / 2
x ≈ 0.00477M (rounded to 5 significant figures)

Now, using the concentration of H+ derived from the NH4+ concentration:

[H+] = 0.00477M

Then, we can calculate the pH as pH = -log[H+]:

pH = -log(0.00477)
pH ≈ 2.32 (rounded to 3 significant figures)

Therefore, the pH of a 0.025M solution of ammonia is approximately 2.32.

Answer 4

suppose many children have shown mild sign of overdose after taking paracetamol at an appropriate and experts recommendation is that the concentration of the drug be reduced to 0.21M (mol/L).Calculate the volume of de-ionized water that could be added to the added to the syrup. Whose initial volume and concentration are indicated on the bottle (in mg/mL), in order to dilute it to the proposed concentration .concentration written on the bottle is 250mg/mL Show all your working.

Answer 5

To calculate the volume of de-ionized water needed to dilute the paracetamol syrup from its initial concentration to a concentration of 0.21M (mol/L), we need to convert the initial concentration from mg/mL to M.

Given:
Initial concentration of the paracetamol syrup: 250 mg/mL
Target concentration: 0.21 M

First, we need to convert the initial concentration from mg/mL to M.

1 mg is equal to 0.001 g, and the molar mass of paracetamol (C8H9NO2) is 151.16 g/mol.

Converting the mg/mL to mol/L:
Initial concentration = (250 mg/mL) * (0.001 g/1 mg) * (1 L/1000 mL) * (1 mol/151.16 g) = 0.00165 mol/L

Now we have the initial concentration in M, and we need to calculate the volume of water needed to achieve the target concentration of 0.21 M.

The dilution equation is:
C1V1 = C2V2

C1 is the initial concentration
V1 is the initial volume
C2 is the target concentration
V2 is the final volume (which is the sum of the initial volume and the volume of water added)

Rearranging the equation for V2, we get:

V2 = (C1 * V1) / C2

Substituting the values, we have:

V2 = (0.00165 mol/L * V1) / 0.21 mol/L

Simplifying, we get:

V2 = 0.007857 * V1

Now, suppose the initial volume (V1) is given on the bottle. Let's assume it is 100 mL.

V2 = 0.007857 * 100 mL
V2 = 0.7857 mL

Therefore, to dilute the paracetamol syrup from its initial concentration of 250 mg/mL to a concentration of 0.21 M, approximately 0.7857 mL of de-ionized water should be added to the syrup.

Please note that this is a theoretical calculation, and in practice, it might be challenging to accurately measure such small volumes.

Answer 6

A person has stung by wasps which produce an alkaline toxic whose concentration is 3M(3moles per cubic decimeter).And there is no acid substance with the same concentration to neutralize the base .But there is the bottle in lab of aqueous hydrochloric acid whose concentration is 98%.describe with the appropriate calculation how you could prepare a 3M(3moles per cubic decimeter ) solution with a volume of 200mL from the stock solution (3M HCL) so that you can use to neutralize the alkaline toxic.

Answer 7

To prepare a 3M hydrochloric acid (HCl) solution with a volume of 200 mL from the stock solution of 98%, we need to calculate the volume of the stock solution and the volume of water required.

Given:
Concentration of stock HCl solution: 98%
Target concentration: 3M
Target volume: 200 mL

First, we need to calculate the volume of the stock solution required to obtain 3 moles of HCl.

Moles of HCl = Molarity * Volume (in L)
3 moles = 3M * Volume (in L)

So, the volume of the stock HCl solution required can be calculated as:
Volume (in L) = 3 moles / 3M

Volume (in L) = 1 L

Since the stock HCl solution is 98% concentration, 1 L of the stock solution contains:
1 L * 0.98 = 0.98 L or 980 mL of HCl.

Now, to prepare a 3M HCl solution with a volume of 200 mL, we need to calculate the volume of water required to dilute the stock solution.

Volume of water = Target volume - Volume of stock HCl solution
Volume of water = 200 mL - 980 mL
Volume of water = -780 mL (negative indicates the volume to be taken out)

Since it is not possible to add negative volume, we realize that we need to adjust the volume of the stock solution used.

To obtain a 3M HCl solution with a volume of 200 mL, we need to reduce the volume of the stock HCl solution.

Given that the stock solution is concentrated at 98%, we can calculate the dilution factor as the ratio of the target concentration to the stock concentration:

Dilution factor = Target concentration / Stock concentration
Dilution factor = 3M / 98%

Now, we can calculate the volume of the stock solution required by dividing the target volume by the dilution factor:

Volume of stock solution = Target volume / Dilution factor
Volume of stock solution = 200 mL / (3M / 98%)

Simplifying, we have:

Volume of stock solution = (200 mL * 98%) / 3M

Volume of stock solution ≈ 6526.67 mL ≈ 6.53 L (rounded to 2 decimal places)

Therefore, to prepare a 3M HCl solution with a volume of 200 mL using the stock solution of 98%, we need to take approximately 6.53 L of the stock solution and then dilute it with water to reach a final volume of 200 mL.

Answer 8

1. Calculate the hydrogen concentration ion for each of the following substances

Blood plasma, pH = 7.4
Orange juice, pH = 3.5
Human Urine, pH = 6.2
Gastric Juice, pH = 1.8
Household ammonia, pH =11.5
2. What is the [𝐶𝐻_3 𝐶𝑂𝑂^− ]/[𝐶𝐻_3 𝐶𝑂𝑂𝐻] ratio in an acetate buffer at pH 5.00
3. How would the following buffers differ in their capacity
0.01M 〖𝑁𝑎〗_2 𝐻𝑃𝑂_4 𝑎𝑛𝑑 0.01𝑀 𝑁𝑎𝐻_2 𝑃𝑂_4
0.1M 〖𝑁𝑎〗_2 𝐻𝑃𝑂_4 𝑎𝑛𝑑 0.1𝑀 𝑁𝑎𝐻_2 𝑃𝑂_4
1.0M 〖𝑁𝑎〗_2 𝐻𝑃𝑂_4 𝑎𝑛𝑑 1.0𝑀 𝑁𝑎𝐻_2 𝑃𝑂_4

Answer 9

1. To calculate the hydrogen ion concentration for each substance given the pH values, we can use the equation pH = -log[H+].

a) Blood plasma, pH = 7.4
[H+] = 10^(-pH) = 10^(-7.4) ≈ 3.98 × 10^(-8) M

b) Orange juice, pH = 3.5
[H+] = 10^(-pH) = 10^(-3.5) = 3.16 × 10^(-4) M

c) Human urine, pH = 6.2
[H+] = 10^(-pH) = 10^(-6.2) ≈ 1.58 × 10^(-7) M

d) Gastric juice, pH = 1.8
[H+] = 10^(-pH) = 10^(-1.8) ≈ 1.58 × 10^(-2) M

e) Household ammonia, pH = 11.5
[H+] = 10^(-pH) = 10^(-11.5) ≈ 3.16 × 10^(-12) M

2. The ratio [CH3COO-] / [CH3COOH] in an acetate buffer can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log([A-] / [HA])

Given that the pH is 5.00, we need to find the pKa value for acetic acid (CH3COOH).

The pKa for acetic acid is approximately 4.76.

pH = 4.76 + log([CH3COO-] / [CH3COOH])

5.00 = 4.76 + log([CH3COO-] / [CH3COOH])

log([CH3COO-] / [CH3COOH]) = 5.00 - 4.76

log([CH3COO-] / [CH3COOH]) ≈ 0.24

Taking the anti-log of both sides:

[CH3COO-] / [CH3COOH] ≈ 10^(0.24)

[CH3COO-] / [CH3COOH] ≈ 1.58

Therefore, the [CH3COO-] / [CH3COOH] ratio in the acetate buffer at pH 5.00 is approximately 1.58.

3. The capacity of a buffer is determined by its concentration and the relative amounts of the acid and conjugate base components.

In this case, we have three buffer systems with different concentrations of Na2HPO4 and NaH2PO4:

a) 0.01 M Na2HPO4 and 0.01 M NaH2PO4
b) 0.1 M Na2HPO4 and 0.1 M NaH2PO4
c) 1.0 M Na2HPO4 and 1.0 M NaH2PO4

The capacity of a buffer is directly proportional to its concentration. Therefore, the buffers with higher concentrations will have a higher capacity to resist changes in pH.

In this case, the buffer with the highest capacity would be the one with 1.0 M Na2HPO4 and 1.0 M NaH2PO4.

Answer 10

. What is the [𝐶𝐻_3 𝐶𝑂𝑂^− ]/[𝐶𝐻_3 𝐶𝑂𝑂𝐻] ratio in an acetate buffer at pH 5.00

3. How would the following buffers differ in their capacity
0.01M 〖𝑁𝑎〗_2 𝐻𝑃𝑂_4 𝑎𝑛𝑑 0.01𝑀 𝑁𝑎𝐻_2 𝑃𝑂_4
0.1M 〖𝑁𝑎〗_2 𝐻𝑃𝑂_4 𝑎𝑛𝑑 0.1𝑀 𝑁𝑎𝐻_2 𝑃𝑂_4
1.0M 〖𝑁𝑎〗_2 𝐻𝑃𝑂_4 𝑎𝑛𝑑 1.0𝑀 𝑁𝑎𝐻_2 𝑃𝑂_4