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makkiz [27]
3 years ago
6

What concentration of nh4no3 is required to make [oh− ] = 1.0 × 10−5 in a 0.200-m solution of nh3?

Chemistry
2 answers:
Evgesh-ka [11]3 years ago
8 0

Answer:

We need a concentration of 0.361M for NH4NO3

Explanation:

Step 1: data given

[OH-] = 1.0 * 10^-5 M

Molarity of the NH3 solution = 0.200 M

Step 2: The balanced equation

NH3 + H2O --> NH4+ + OH-

Step 3: Calculate pOH

pH = pKa + log [A-]/[HA]

pOH = pKb + log [B+]/[BOH]

⇒ B+ = base's conjugate acid

⇒ BOH = the base

if [OH-] = 1.0 * 10^-5 M

then pOH = -log ( 1.0 * 10^-5)

pOH = 5

Step 4: Calculate [NH4+]

Kb of NH3 = 1.8 *10^-5

pKb = -log (1.8 * 10^-5) =  4.744

pOH = pKb + log [B+]/[BOH]

⇒pOH = 5

⇒pKb = 4.744

⇒[B+] = [NH4+] = TO BE DETERMINED

⇒[BOH] = [NH3] = 0.200 M

5 - 4.744 = log [NH4+]/[0.200]

0.256 = log [NH4+]/[0.200]

10^0.256 = [NH4+]/[0.200]

1.80 = [NH4+]/[0.200]

0.361 = [NH4+]

Step 5: Calculate [NH4NO3]

NH4NO3 --> NH4+ + NO3-

[NH4NO3] = 0.361 M

We need a concentration of 0.361M for NH4NO3

nikdorinn [45]3 years ago
7 0

Answer:

The concentration of NH₄NO₃ required to make [OH⁻]  = 1.0 × 10⁻⁵ in a 0.200-m solution of NH₃ is 0.3639 M

Explanation:

Here we have

The reaction of Ammonia and water to produce OH⁻

NH₃ + H₂O → NH₄⁺ + OH⁻

From Henderson-Hasselbalch Equation for bases we have

pOH = pKb + log[acid]/[base]

The concentration of OH⁻ is given as [OH⁻] = 1.0 × 10⁻⁵

Therefore pOH = -log [OH⁻]  = -log(1.0 × 10⁻⁵) = 5

The base dissociation constant is given by  

kb = [B⁺][OH⁻]/[BOH]

For NH₃, kb = 1.8×10⁻⁵

pKb = -log Kb = 4.74

From the Henderson-Hasselbalch Equation

The acid concentration  = [NH₄⁺] and the base concentration = [NH₃] = 0.200 M

pOH = pKb + log ([acid]/[base])

Substituting the values, we have

5 = 4.74 + log ([ NH4+ ]/0.200 M)

0.26=log\frac{[NH_4^+]}{0.200 \hspace {0.09cm}M}

10^{0.26} =1.8197=\frac{[NH_4^+]}{0.200 \hspace {0.09cm}M}

[NH4+] = 1.8197×0.200 M = 0.3639 M

NH₄NO₃ → NH₄⁺ + NO₃⁻

Therefore, since one mole of NH₄NO₃, produces one mole of NH₄⁺, the number of moles of NH₄NO₃ required to produce  0.3639 M of NH₄⁺ is 0.3639 M of NH₄NO₃

The concentration of NH₄NO₃ required to make [OH⁻]  = 1.0 × 10⁻⁵ in a 0.200-m solution of NH₃ = 0.3639 M

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How many mol of C7H16 would you have if you have 76.36 grams? Give your answer to 2 decimal spaces.
tensa zangetsu [6.8K]

0.761 mol of C_7H_{16}  would you have if you have 76.36 grams.

<h3>What is a mole?</h3>

A mole is defined as 6.02214076 × 10^{23}of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Given data:

Mass=76.36 grams

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What is an atom? A- the smallest particle of an element. B- the negatively charged particle found in a cloud. C- the densely pac
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An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni
Svetllana [295]

<u>Answer:</u> The pH of acid solution is 4.58

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}    .....(1)

  • <u>For KOH:</u>

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol

  • <u>For propanoic acid:</u>

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol

The chemical reaction for propanoic acid and KOH follows the equation:

                 C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O

<u>Initial:</u>          0.1372         0.04514  

<u>Final:</u>           0.09206          -                0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L     (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})

pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})

We are given:  

pK_a = negative logarithm of acid dissociation constant of propanoic acid = 4.89

[C_2H_5COOK]=\frac{0.04514}{0.26594}

[C_2H_5COOH]=\frac{0.09206}{0.26594}

pH = ?  

Putting values in above equation, we get:

pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58

Hence, the pH of acid solution is 4.58

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3 years ago
Molar mass in sulfuric acid
EleoNora [17]

Hey there!

Sulfuric acid: H₂SO₄

H: 2 x 1.008 = 2.016

S: 1 x 32.065 = 32.065

O: 4 x 16 = 64

--------------------------------

                 98.081 g/mol

The molar mass of sulfuric acid is 98.081 g/mol.

Hope this helps!

8 0
3 years ago
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