Science equipment is drawn in two dimensions

This is because a buffer has the ability to not get affected by the addition of small amounts of an acid or a base. So, basically it keeps the concentration of both hydrogen ions and hydroxides equal. As a result, it helps in maintaining the pH of the solution.

And, the capacity of a buffer solution to resist the change is known as buffer capacity.

Thus, we can conclude that buffering capacity refers to the extent to which a buffer solution can counteract the effect of added acid or base.

8 0

3 years ago

Which element begins with the ketter K?

torisob [31]

Krypton is the only element that begins with the letter K.

4 0

3 years ago

Consider the titration of 40.0 mL of 0.200 M HClO4 by 0.100 M KOH. Calculate the pH of the resulting solution after the followin

olya-2409 [2.1K]

Answer:

a) 0.70

b) 7.00

c) 0.85

d) 12.15

e) 1.30

Explanation:

The neutralization reaction involved in the titration is:

HClO₄(aq) + KOH(aq) → KClO₄(aq) + H₂O(l)

According to the chemical equation, 1 mol of HClO₄ reacts with 1 mol of KOH (1 equivalent of acid with 1 equivalent of base). The moles are calculated from the product of the molar concentration (M) and the volume in liters.

We have the following moles of acid (HClO₄):

40.0 mL x 1 L/1000 mL = 0.04 L

0.200 mol/L x 0.04 L = 8 x 10⁻³ moles HClO₄

Since HClO₄ is a strong acid (completely dissociated into H⁺ and ClO₄⁻ ions), the moles of HClO₄ are equal to the moles of H⁺. Then, we can calculate the initial pH:

[H⁺] = 0.200 M → pH = -log [H⁺] = -log (0.200) = 0.70

Now, we calculate the pH after the addition of KOH. Since KOH is a strong base, the concentration of KOH is equal to the concentration of OH⁻ ions.

a) 0.0 mL

No KOH is added, so the pH is the initial pH: 0.70

b) 80.0 mL KOH

80.0 mL x 1 L/1000 mL = 0.08 L

0.100 mol/L x 0.08 L = 8 x 10⁻³ moles KOH = 8 x 10⁻³ moles OH⁻

After neutralization we have:

8 x 10⁻³ moles H⁺ - 8 x 10⁻³ moles OH⁻ = 0

The neutralization reaction is complete and there is no remaining H⁺ from the acid. The concentration of H⁺ is equal to the concentration of H⁺ of water:

[H⁺] = 1 x 10⁻⁷ M → pH = -log [H⁺] = -log (1 x 10⁻⁷) = 7.0

c) 10.0 mL KOH

10.0 mL x 1 L/1000 mL = 0.01 L

0.100 mol/L x 0.01 L = 1 x 10⁻³ moles KOH = 1 x 10⁻³ moles OH⁻

After neutralization we have:

8 x 10⁻³ moles H⁺ - 1 x 10⁻³ moles OH⁻ = 7 x 10⁻³ moles H⁺

The total volume is: V = 40.0 mL + 10.0 mL = 50 mL = 0.05 L

[H⁺] = 7 x 10⁻³ moles/0.05 L = 0.14 → pH = -log [H⁺] = -log (0.14) = 0.85

d) 100.0 mL KOH

100.0 mL x 1 L/1000 mL = 0.1 L

0.100 mol/L x 0.1 L = 0.01 moles KOH = 1 x 10⁻² moles OH⁻

After neutralization we have:

1 x 10⁻² moles OH⁻ - 8 x 10⁻³ moles H⁺ = 2 x 10⁻³ moles OH⁻

The total volume is: V = 40.0 mL + 100.0 mL = 140 mL = 0.14 L

[OH⁻] = 2 x 10⁻³ moles/0.14 L = 0.014 → pOH = -log [OH⁻] = -log (0.014) = 1.84

pH + pOH = 14 → pH = 14 - pOH = 14 - 1.84 = 12.15

e) 40.0 mL KOH

40.0 mL x 1 L/1000 mL = 0.04 L

0.100 mol/L x 0.04 L = 4 x 10⁻³ moles KOH = 4 x 10⁻³ moles OH⁻

After neutralization we have:

8 x 10⁻³ moles H⁺ - 4 x 10⁻³ moles OH⁻ = 4 x 10⁻³ moles H⁺

The total volume is: V = 40.0 mL + 40.0 mL = 80.0 mL = 0.08 L

[OH⁻] = 4 x 10⁻³ moles/0.08 L = 0.05 M → pH = -log [H⁺] = -log (0.05) = 1.30

5 0

3 years ago