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aivan3 [116]
1 year ago
13

Calculate the ph of a solution formed by mixing 250. 0 ml of 0. 15 m nh4cl with 200. 0 ml of 0. 12 m nh3. the kb for nh3 is 1. 8

× 10-5.
i. 9. 45
ii. 4. 74
iii. 9. 06
iv. 04. 55
v. 9. 26
Chemistry
1 answer:
Phantasy [73]1 year ago
4 0

The potential of the hydrogen in water gives the pH of the acidity and basicity of the solution. The pH of the solution is 9.06. Thus, option iii is correct.

<h3>What is pH?</h3>

The pH of a substance is given by the subtraction of the pOH from 14 which is the range of the pH scale.

The dissociation reaction for ammonia is given as,

NH₃ (aq) + H₂O (l)  ⇄  NH₄⁺ (aq) + OH⁻ (aq)

Here, the concentration of ammonia is [NH₃] - x, ammonium ion is [NH₄⁺] + x, and hydroxide ion is x.

The molar concentration of ammonia is,

M = (0.12 M × 0.2 L) ÷ 0.45 L = 0.053 M

The molar concentration of ammonium ion is,

M = (0.15 M × 0.25 L) ÷ 0.45 L = 0.083 M

From the base dissociation constant and previous concentration from the reaction, the value of x or hydroxide ion is calculated as,

Kb = [NH₄⁺][OH⁻] ÷ [NH₃]

1.8 × 10⁻⁵ (0.053 - x) - (0.083 + x) × x = 0

x = [OH⁻] = 1.15 x 10⁻⁵

The pH from the hydroxide ion is calculated as,

pOH = - log [OH⁻]

= - log (1.15 x 10⁻⁵)

= 4.94

Further,

pH = 14 - pOH

= 14 - 4.94

= 9.06

Therefore, option iii. 9.06 is the pH of the solution.

Learn more about pH here:

brainly.com/question/14790520

#SPJ4

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Answer:Sample Absorbance (625 nm)  

A 0.536  

B 0.783  

C 0.045  

Therefore, I will use these data to solve your question. If you have other absorbances values, just follow my steps and plug in different numbers.

First, we see 1 mole of NH3 gives 1 mole product.

In B moles of NH3 = moles of NH3 in A + (5.5 x10^-4 x2.5/1000) = 1.375 x10^6 + mA

( mA = moles of NH3 in A) vol of B = 25 = vol of A

now A = el C = eC ( since l = 1cm)

Because, n net absorbance due to complex blank absorbance must be removed.

Here A(A) = 0.536 - 0.045 = 0.491 , A(B) = 0.783 - 0.045 = 0.738  

(you can plug in different numbers in this step)

A2/A1 = C2/C1 , A(B)/A(A) = (1.375x10^-6 +mA)/(mA) = 0.738/0.491

So, mA = 2.733 x 10^-6 = moles of NH3 in A (Lake water)

Hence [NH3] water ( 2.733 x10^-6 ) x 1000/25 = 1.093 x 10^-4 M

Lake water vol = 10 ml out of 25,

Concentration of ammonia in lake water = 2.733 x10^-6 x 1000/10 = 2.733 x 10^-4 M

Then, A = 0.491 = e x 1 x 1.093 x10^-4

e = 4492 M-1cm-1

Explanation:

4 0
3 years ago
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