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alexandr402 [8]
3 years ago
7

3. Determine the pH of each of the following solutions. (Hint: See Sample Problem B.) a. 1.0 x 10-2 M HCI c. 1.0 x 10-MHI b. 1.0

x 10-3 M HNO3 d. 1.0 x 10-M HB​
Chemistry
1 answer:
densk [106]3 years ago
3 0

Answer:

a. pH = 2 b. pH = 3 c. pH = 1 d. Unanswerable

Explanation:

pH = -log[H+] OR pH = -log{H3O+]

                    and inversely

pOH = -log[OH-]

1. Determine what substance you are working with, (acid/base)

2. Determine whether or not that acid or base is strong or weak.

a. 1.0 x 10^-2M HCl

HCl is a strong acid, therefore it will dissociate completely into H+ and Cl- with all ions going to the H+, therefore, the concentration of HCl and concentration of H+ are going to be equal, meaning we simply take the negative logarithm of the concentration of HCl and that would equal pH

pH = -log[H+]

pH = -log(1.0x10^-2)

pH = 2

b. 1.0 x 10^-3M HNO3

HNO3 like part a, is a strong acid, therefore it would simply require you to take the negative logarithm of the concentration of the compound itself, to find its pH.

pH = -log[H+]

pH = -log(1.0 x 10^-3)

pH = 3

c. 1.0 x 10^-1M HI

Like the previous parts, HI is a strong acid

pH = -log[H+]

pH = -log(0.10)

pH = 1

d. HB isn't an element, nor is it a compound so that would be unanswerable.

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a certain compound was found to contain 54.0 g of carbon and 10.5 grams of hydrogen. its relative molecular mass is 86.0. find t
dexar [7]

Answer:

empirical formula = C3H7

molecular formula = C6H14

3 0
3 years ago
A solution of 0.0470 M HCl is used to titrate 26.0 mL of an ammonia solution of unknown concentration. The equivalence point is
DanielleElmas [232]

The pH at equivalence point is 12.46

At equivalence point, number of moles of acid, n equals number of moles of base, n'

So, n = n'

CV = C'V' where

  • C = concentration of acid (HCl) = 0.0470 M,
  • V = volume of acid = 16.0 mL,
  • C' = concentration of base (ammonia solution) and
  • V' = volume of base = 26.0 mL.
<h3>Concentration of ammonia solution</h3>

Making C' subject of the formula, we have

C' = CV/V'

Substituting the values of the variables into the equation, we have

C' = CV/V'

C' = 0.0470 M × 16.0 mL/26.0 mL

C' = 0.752 MmL/26.0 mL

C' = 0.0289 M

<h3>The concentration of acid at equivalence point</h3>

We know that the ion-product of water Kw is

Kw = [H⁺][OH⁻] =  where

  • [H⁺] = concentration of HCl at equivalence point,
  • [OH⁻] = C' = concentration of ammonia solution = 0.0289 M and
  • Kw = 1.01 × 10⁻¹⁴

Making [H⁺] subject of the formula, we have

[H⁺} = Kw/[OH⁻]

[H⁺] = 1.01 × 10⁻¹⁴/0.0289

[H⁺] = 34.95 × 10⁻¹⁴

[H⁺] = 3.495 × 10⁻¹³

<h3>pH at equivalence point</h3>

Since pH = -㏒[H⁺]

pH = -㏒[3.495 × 10⁻¹³]

pH = -㏒[3.495] + (-㏒10⁻¹³)

pH = -㏒[3.495] + [-13(-㏒10)]

pH = 13 - 0.5434

pH = 12.4566

pH ≅ 12.46

So, the pH at equivalence point is 12.46

Learn more about pH at equivalence point here:

brainly.com/question/25487920

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