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kari74 [83]
3 years ago
10

For each solution, calculate the initial and final pH after the addition of 0.010 mol of NaOH.

Chemistry
1 answer:
Murljashka [212]3 years ago
8 0

Answer:

A. 12.6

B. 4.05

C. 10.93

Explanation:

A. 0.010mol of NaOH in 250.0mL gives a concentration of 0.04M NaOH = 0.04M OH⁻

pOH = -log [OH⁻] = 1.398

pH = 14-pOH = <em>12.6</em>

<em></em>

B. The reaction of NaOH with HCHO₂ is:

NaOH + HCHO₂ → H₂O + CHO₂⁻ + Na⁺

Initial moles of CHO₂⁻ and HCHO₂ are:

CHO₂⁻ = 0.250L × (0.275mol/L) = 0.06875moles

HCHO₂ = 0.250L × (0.195mol/L) = 0.04875moles

After reaction:

CHO₂⁻ = 0.06875moles + 0.010mol = 0.07875mol

HCHO₂ = 0.04875moles - 0.010mol = 0.03875mol

Using H-H equation (pKa of this buffer: 3.74)

pH = 3.74 + log [CHO₂⁻] / [HCHO₂]

pH =  3.74 + log [0.07875mol] / [0.03875mol]

<em>pH = 4.05</em>

<em></em>

C. The reaction of NaOH with CH₃CH₂NH₃Cl is:

NaOH + CH₃CH₂NH₃Cl → H₂O + CH₃CH₂NH₂ + Cl⁻ + Na⁺

Initial moles of CH₃CH₂NH₃Cl and CH₃CH₂NH₂ are:

CH₃CH₂NH₃Cl = 0.250L × (0.235mol/L) = 0.05875moles

CH₃CH₂NH₂ = 0.250L × (0.255mol/L) = 0.06375moles

After reaction:

CH₃CH₂NH₃Cl = 0.05875moles - 0.010mol = 0.04875mol

CH₃CH₂NH₂ = 0.06375moles + 0.010mol = 0.07375mol

Using H-H equation (pKa of this buffer: 10.75)

pH = 10.75 + log [CH₃CH₂NH₂] / [CH₃CH₂NH₃Cl]

pH =  10.75 + log [0.07375mol] / [0.04875mol]

<em>pH = 10.93</em>

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3 years ago
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Answer:

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Explanation:

You just need to first get the atomic weights of the elements involved. You can easily get these from your periodic table.

If you are going to do this properly, please use the weight with at least two decimal places for accuracy (e.g. 15.99 g/mol).

Also, please take note that I will be using the unit g/mol for all the weights. Thus,

Step 1

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Since your compound is  

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N

H

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)

2

C

O

3

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Step 2

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H = 1.008 g/mol × (4×2) =

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O = 16.00 g/mol × 3 =

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C = 12.01 g/mol × 1 =

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To get the mass of the substance, we need to add all the weights from Step 2.

Step 3

molar mass of

(

NH

4

)

2

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3

=

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=

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