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il63 [147K]
3 years ago
15

A buffer is prepared by dissolving 0.80 moles of NH3 and 0.80 moles of NH4Cl in 1.00 L of aqueous solution. If 0.10 mol of NaOH

is added to 250 mL of this buffer, what is the pH of the resultant solution?
a.10.54
b.4.27
c.8.78
d.9.73
e.5.22
Chemistry
1 answer:
Vikki [24]3 years ago
7 0

Answer:

a. 10.54

Explanation:

reaction is

NH₃ +  NaOH   -----------------NH₄Cl + H₂O

0.10 mol NaOH will consume 0.10 mol NH₃ thereby decreasing the initial amount of moles NH₃ and increasing that of NH₄Cl

mol NH₃  = 0.80 - 010 = .70

mol NH₄Cl = 0.80 + .10 = 0.90

pH = pKₐ + log (( NH₃/NH₄Cl))

pH =  9.26 + log (( 0.90 + 0.70)) = 9.26 + 0.11 = 10.54

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What is the concentration, in m/v percent, of a solution prepared from 50 g NaCl and 2.5 L of solution?
Serjik [45]
When we convert the given mass in grams and volume in liters to m/v percent, we recall that m/v percent is expressed as grams/100 milliliters. In this case the expression becomes (50 grams/ 2500 L)*(0.1L/100ml), that is equal to 0.002 grams/ 100 mL. Hence the the concentration is equal to 0.2 m/v percent.
4 0
3 years ago
The ka of hypochlorous acid (hclo) is 3.0 x 10-8 at . what is the % ionization of hypochlorous acid in a 0.015 m aqueous solutio
Ahat [919]

Answer is: the % ionization of hypochlorous acid is 0.14.

Balanced chemical reaction (dissociation) of an aqueous solution of hypochlorous acid:

HClO(aq) ⇄ H⁺(aq) + ClO⁻(aq).

Ka = [H⁺] · [ClO⁻] / [HClO].

[H⁺] is equilibrium concentration of hydrogen cations or protons.

[ClO⁻] is equilibrium concentration of hypochlorite anions.

[HClO] is equilibrium concentration  of hypochlorous acid.

Ka is the acid dissociation constant.

Ka(HClO) = 3.0·10⁻⁸.

c(HClO) = 0.015 M.

Ka(HClO) = α² · c(HClO).

α = √(3.0·10⁻⁸ ÷ 0.015).

α = 0.0014 · 100% = 0.14%.

5 0
3 years ago
If you have 400 grams of a substance that decays with a half-life of 14 days, then how much will you have after 56 days? To help
alisha [4.7K]

Answer:

25 grams

Explanation:

You strat off with 400 grams of your substance. By day 14, half has dcayed and you only have 200 grams left. By day 28, there are 100 grams of the substance. On day 42, there are 50 grams left. Finally, on day 56, the substance has been through four half-lives and 25 grams remain.

3 0
2 years ago
Easy question, need help
jeka94

Answer: B= 210 amps

Explanation:

175

0.75 so you divide it so the answer it 210

since first you find for one

then you multiply 0.75 to the answer you get

Hope this helps :)

8 0
2 years ago
Read 2 more answers
What is the majority of 0.50 g of na dissolved in a 1.5 L solution
Oxana [17]

The correct answer is 0.014467 M.  

Molarity is defined as the number of moles present in a liter solution, that is, number of moles / liter solution.  

The molar mass of sodium (Na) is 23.0 g/mol

Thus, 1 mole of Na contains 23.0 g

Now, x moles of Na contains 0.50 g

Moles = 0.50 × 1 / 23.0

Moles = 0.50 / 23.0  

= 0.0217 moles of Na

Molarity = Number of moles / liters of solution

= 0.0217 / 1.5

= 0.014467 M


8 0
3 years ago
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