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aev [14]
3 years ago
11

For each of the following compounds, indicate the pH at which 50% of the compound will be in a form that possesses a charge and

at which pH more than 99% of the compound will be in a form that possesses a charge.
ClCH2COOH (pKa = 2.86)
CH3CH2NH+3 (pKa = 10.7)

Express your answer using two decimal places

a. Determine a pH at which 50% of ClCH2COOH will be in a form that possesses a charge.
b. Determine a pH at which pH more than 99% of ClCH2COOH will be in a form that possesses a charge.
c. Determine a pH at which 50% of CH3CH2NH+3 will be in a form that possesses a charge.
d. Determine a pH at which pH more than 99% of CH3CH2NH+3 will be in a form that possesses a charge.
Chemistry
1 answer:
Virty [35]3 years ago
5 0

Answer:

a. 2..86 b. 4.86 c. 10.7 d. 8.7

Explanation:

a. Determine a pH at which 50% of ClCH2COOH will be in a form that possesses a charge.

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA]

where [A⁻] = concentration of conjugate base (or charged form) and [HA] = concentration of acid.

At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1

So, pH = pKa + log[A⁻]/[HA]

pH = pKa + log1

pH = pKa = 2.86

b. Determine a pH at which pH more than 99% of ClCH2COOH will be in a form that possesses a charge.

Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.99x while the acidic concentration is remaining 1 % (1 - 0.99)x = 0.01x

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base (or charged form) = 0.99x and [HA] = concentration of acid = 0.01x.

pH = pKa + log0.99x/0.01x

pH = pKa + log0.99/0.01

pH = 2.86 + log99

pH = 2.86 + 1.996

pH = 4.856

pH ≅ 4.86

c. Determine a pH at which 50% of CH3CH2NH+3 will be in a form that possesses a charge.

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA]

where [A⁻] = concentration of conjugate base and [HA] = concentration of acid.

At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1

So, pH = pKa + log[A⁻]/[HA]

pH = pKa + log1

pH = pKa = 10.7

d. Determine a pH at which pH more than 99% of CH3CH2NH+3 will be in a form that possesses a charge.

Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.01x while the acidic concentration is remaining 99 % (1 - 0.01)x = 0.99x (which possesses the charge).

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base = 0.01x and [HA] = concentration of acid = 0.99x.

pH = pKa + log0.01x/0.99x

pH = pKa + log1/99

pH = 10.7 - log99

pH = 10.7 - 1.996

pH = 8.704

pH ≅ 8.7

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5 0
3 years ago
The H⁺ concentration in an aqueous solution at 25 °C is 9.1 × 10⁻⁴. What is [OH⁻]?
Vinvika [58]

Steps:-

  • First we calculate pH then pOH then [OH-]

\\ \tt\rightarrowtail pH=-log[H^+]

\\ \tt\rightarrowtail pH=-log[9.1\times 10^{-4}]

\\ \tt\rightarrowtail pH=-log9.1-log10^{-4})

\\ \tt\rightarrowtail pH=0.95+4

\\ \tt\rightarrowtail pH=4.95

Now

\\ \tt\rightarrowtail pH+pOH=14

\\ \tt\rightarrowtail pOH=14-4.95

\\ \tt\rightarrowtail pOH=9.05

So

\\ \tt\rightarrowtail -log[OH^-]=9.05

\\ \tt\rightarrowtail log[OH^-]=-9.05

\\ \tt\rightarrowtail OH^-=10^{-9.05}

\\ \tt\rightarrowtail OH^-=8.91\times 10^{-4}

5 0
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How many moles are in 1.5 x 10^20 molecules of NH3?
Keith_Richards [23]
<span>Because the question is asking moles of NH3, the compound, any subscripts are irrelevant. It only wants to know how many moles of NH3 molecules, not individual atoms.

Therefore, we can simply convert to moles. 1.5x10^23/6.022x10^23 = .249 moles of NH3.

(If it were to ask moles of Hydrogen, for example, you would multiply the answer by 3, because there are 3 atoms of Hydrogen per one molecule of NH3. But this only asks for moles of the entire compound).  

hope you have a great day! :)
</span>
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7 0
3 years ago
A solution was prepared by mixing 20.00 mL of 0.100 M and 120.00 mL of 0.200 M. Calculate the molarity of the final solution of
Marat540 [252]

Answer:

0.186M

Explanation:

First, we need to obtain the moles of nitric acid that are given for each solution. Then, we need to divide these moles in total volume (120mL + 20mL = 140mL = 0.140L) to obtain molarity:

<em>Moles Nitric acid:</em>

0.0200L * (0.100mol / L) = 0.00200 moles

0.120L * (0.200mol / L)= 0.02400 moles

Total moles: 0.02400moles + 0.00200moles = 0.026 moles of nitric acid

Molarity: 0.026 moles / 0.140L

<h3>0.186M</h3>
6 0
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