Is limestone a reactant or a product

Find the pH of the equivalence point(s) and the volume (mL) of 0.0372 M NaOH needed to reach the point(s) in titrations of(a) 42

Neko [114]

The pH of the equivalence points is 8.54

Concept of pH

A solution's acidity or alkalinity can be determined based on the concentration of hydrogen ions in the solution, or pH. Acidic aqueous solutions at 25 °C have a pH under 7, while basic or alkaline aqueous solutions have a pH above 7. Since the concentration of H3O+ is equal to the concentration of OH in pure water, a pH level of 7.0 at 25°C is referred to as "neutral". Strong bases may have a pH above 14, while very strong acids may have a pH below 14.

0.0520 M CH3COOH in 42.2 mL of moles is as follows:

2.194x103 mol CH3COOH = 0.0422L (0.0520mol / L)

that react with NaOH, resulting in:

NaOH + CH3COOH = CH3COO + Na+ + H2O

Thus, 1 mole of acetic acid and 1 mole of NaOH react.

As a result, 2.194x103 mol of NaOH are required to reach the equivalence point in volume:

To attain the equivalency point, 2.194x103 mol (1L / 0.0372mol) = 0.05899L 58.99mL of 0.0372 M NaOH

You will only have CH3COO at the equivalency point because it is in equilibrium with water, so:

H2O(l) + CH3COO(aq) CH3COOH(aq) + OH (aq)

A definition of equilibrium is:

Kb = 5.6x1010 = [OH] / [CH3COO] / [CH3COOH]

2.194x103mol of CH3COO has a molarity of (0.05899L + 0.0422L) = 0.02168M.

Therefore, equilibrium concentrations are:

[CH3COO]=0.02168M-X [CH3COOH]=X [OH]=X

5.6x1010 = [X] [X] / [0.02168M - X] converts to Kb.

1.214x1011 - 5.6x1010X = X2 X2 + 5.6x1010X - 1.214x1011 = 0 Finding the value of X:

False response; there are no negative concentrations. X: -3.48x106

As [OH] = X, [OH] = 3.484x106M, X is 3.484x106.

As 14 = pOH + pH pH = 8.54, pOH = -log [OH], or 5.46.

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3 0

2 years ago

Answer the following questions about the solubility of AgCl(s). The value of Ksp for AgCl(s) is 1.8 × 10−10.

Firlakuza [10]

Answer:

[Ag⁺] = 1.3 × 10⁻⁵M

s = 3.3 × 10⁻¹⁰ M

Because the common ion effect.

Explanation:

1. Value of [Ag⁺] in a saturated solution of AgCl in distilled water.

The value of [Ag⁺] in a saturated solution of AgCl in distilled water is calculated by the dissolution reaction:

AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)

The ICE (initial, change, equilibrium) table is:

AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)

I X 0 0

C -s +s +s

E X - s s s

Since s is very small, X - s is practically equal to X and is a constant, due to which the concentration of the solids do not appear in the Ksp equation.

Thus, the Ksp equation is:

Ksp = [Ag⁺] [Cl⁻]

Ksp = s × s

Ksp = s²

By substitution:

1.8 × 10⁻¹⁰ = s²

s = 1.34 × 10⁻⁵M

Rounding to two significant figures:

[Ag⁺] = 1.3 × 10⁻⁵M ← answer

2. Molar solubility of AgCl(s) in seawater

Since, the conentration of Cl⁻ in seawater is 0.54 M you must introduce this as the initial concentration in the ECE table.

The new ICE table will be:

AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)

I X 0 0.54

C -s +s +s

E X - s s s + 0.54

The new equation for the Ksp equation will be:

Ksp = [Ag⁺] [Cl⁻]

Ksp = s × ( s + 0.54)

Ksp = s² + 0.54s

By substitution:

1.8 × 10⁻¹⁰ = s² + 0.54s

s² + 0.54s - 1.8 × 10⁻¹⁰ = 0

Now you must solve a quadratic equation.

Use the quadratic formula:

$s=\dfrac{-0.54\pm\sqrt{0.54^2-4(1)(-1.8\times 10^{-10})}}{2(1)}$

The positive and valid solution is s = 3.3×10⁻¹⁰ M ← answer

3. Why is AgCl(s) less soluble in seawater than in distilled water.

AgCl(s) is less soluble in seawater than in distilled water because there are some Cl⁻ ions is seawater which shift the equilibrium to the left.

This is known as the common ion effect.

By LeChatelier's principle, you know that an increase in the concentrations of one of the substances that participate in the equilibrium displaces the reaction to the direction that minimizes this efect.

In the case of solubility reactions, this is known as the common ion effect: when the solution contains one of the ions that is formed by the solid reactant, the reaction will proceed in less proportion, i.e. less reactant can be dissolved.

3 0

3 years ago

Assuming all the nh3 dissolves and that the volume of the solution remains at 0.300 l , calculate the ph of the resulting soluti

barxatty [35]

Assume p=735 Torr V= 7.6L R=62.4 T= 295 PV-nRT (735 Torr)(7.60L)= n (62.4Torr-Litres/mole-K)(295K) 0.30346 moles of NH3 Find moles 0.300L solution of 0.300 M HCL = 0.120 moles of HCL 0.30346 moles of NH3 reacts with 0.120 moles of HCL producing 0.120 moles of NH4+ ION, and leaving 0.18346 mole sof NH3 behind Find molarity 0.120 moles of NH4+/0.300L = 0.400 M NH4+ 0.18346 moles of NH3/0.300L = 0.6115 M NH3 NH4OH --> NH4 & OH- Kb = [NH4+][OH]/[NH4OH] 1.8 e-5=[0.300][OH-]/[0.6115] [OH-]=1.6e-5 pOH= 4.79 PH=9.21 .

3 0

3 years ago

According to the selection, what is the main weakness of the tortoise?

Agata [3.3K]

Answer:

Nose

Explanation:

Because every time I touch it's nose, it will hide itself into the shell faster than other places

8 0

3 years ago

A cylinder contains 250 g of Helium at 200 K. The external pressure is constant at 1 atm. The temperature of the system is raise

Black_prince [1.1K]

Answer:

There is 96200 J or 96 kJ of heat released.

Explanation:

Step 1: Data given

Mass of helium = 250 grams

Temperature = 200 K

Temperature is raised by 74 K

The heat capacity of helium = 20.8 J/mol*K

Molecular weight for helium = 4 g/mol

Step 2: Calculate moles of Helium

Moles of Helium = mass of Helium / molar mass of helium

Moles of helium = 250 grams / 4g/mol

Moles of helium = 62.5 moles

Step 3: Calculate heat

Q = n*c*ΔT or

⇒ with Q = the heat released in Joule

⇒ with n = moles of helium = 62.5 moles

⇒ with c = the heat capacity of helium = 20.8 J/mol*K

⇒ with ΔT = the change in temperature = T2 - T1 = 74 K

Q = 62.5 mol * 20.8 J/mol*K * 74 K

Q = 96200 J = 96 kJ

Since the temperature is raised, this is an exothermic reaction. The heat is released.

There is 96200 J or 96 kJ of heat released.

8 0

3 years ago