All categories

2 years ago

A. What is the pH of 8.9 × 10−3 M HCl? B. What is the pH of 8 × 10−8 M HCl?

Chemistry

1 answer:

Leya [2.2K]2 years ago

7 0

Answer:

A. p(H) = 2.0506

B. p(H) = 7.0969

Explanation:

p(H)= -log[H+]

p(H)= -log(8.9×10^-3)

p(H) = 2.0506

p(H)= -log[H+]

p(H)= -log(8×10^-8)

p(H) = 7.0969

You might be interested in

Please write balanced equations for these reactions! WILL GIVE BRAINLIEST! 50 POINTS

Romashka [77]

Answer:

(NH4)2S(aq) + Pb(NO3)2(aq) --> 2NH4NO3 (aq) + PbS (s)

3 0

2 years ago

What most likely happens during this reaction

Bad White [126]

Answer:

I think that it is A I am sorry if I am wrong

Explanation:

8 0

2 years ago

The rate of a certain reaction is given by the following rate law: rate Use this information to answer the question below. At a

AleksAgata [21]

Answer:

Initial rate of the reaction when concentration of hydrogen gas is doubled will be $3.2\times 10^6 M/s$ .

Explanation:

$N_2+3H_2\rightarrow 2NH_3$

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

Initial rate of the reaction = R = $4.0\times 10^5 M/s$

$R = k\times [N_2][H_2]^3$

$4.0\times 10^5 M/s=k\times [N_2][H_2]^3$

The initial rate of the reaction when concentration of hydrogen gas is doubled : R'

$[H_2]'=2[H_2]$

$R'=k\times [N_2][H_2]'^3=k\times [N_2][2H_2]^3$

$R'=8\times k\times [N_2][H_2]^3$

$R'=8\times R=8\times 4\times 10^5 M/s=3.2\times 10^6 M/s$

Initial rate of the reaction when concentration of hydrogen gas is doubled will be $3.2\times 10^6 M/s$ .

7 0

3 years ago

Can you check this for me?

Aliun [14]

Hi, your answer is correct.

3 0

3 years ago

You need to produce a buffer solution that has a pH of 5.50. You already have a solution that contains 10 mmol (millimoles) of a

balandron [24]

Answer:

56.9 mmoles of acetate are required in this buffer

Explanation:

To solve this, we can think in the Henderson Hasselbach equation:

pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])

To make the buffer we know:

CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka

We know that Ka from acetic acid is: 1.8×10⁻⁵

pKa = - log Ka

pKa = 4.74

We replace data:

5.5 = 4.74 + log ([acetate] / 10 mmol)

5.5 - 4.74 = log ([acetate] / 10 mmol)

0.755 = log ([acetate] / 10 mmol)

10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)

5.69 = ([acetate] / 10 mmol)

5.69 . 10 = [acetate] → 56.9 mmoles

6 0

2 years ago