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lorasvet [3.4K]
3 years ago
5

An amphoteric salt is one that contains an anion that can act as either an acid or a base in water. Baking soda, NaHCO3, is an e

xample. By combining the ionization and hydrolysis reactions of the anion, you get the principle reaction that occurs when this salt is dissolved in water.
2HCO3-(aq) CO32-(aq) + H2CO3(aq)
The pH for such, a solution is given by
pH = pKa1 + pKa2/2
where Ka1 and Ka2 are the ionization constants of the acid (in this case, H2CO3). Note that the pH of the solution is independent of the salt concentration. Calculate the pH of a NaHCO3 solution.
Chemistry
1 answer:
LenKa [72]3 years ago
3 0

Answer:

pH = 8.34

Explanation:

The equilbriums of the amphoteric HCO₃⁻ (Ion of NaHCO₃) are:

H₂CO₃ ⇄ <em>HCO₃⁻</em> + H⁺ Ka1 <em>-Here, HCO₃⁻ is acting as a base-</em>

<em>HCO₃⁻</em>⇄ CO₃²⁻ + H⁺ Ka2 <em>-Here, is acting as an acid-</em>

Where Ka1 = 4.3x10⁻⁷ and Ka2 = 4.8x10⁻¹¹. As pKa = -log Ka:

pKa1 = 6.37; pKa2 = 10.32

As the pH of amphoteric salts is:

pH = (pKa1 + pKa2) / 2

<h2>pH = 8.34</h2>
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) determine the theoretical yield and the percent yield if 21.8 g of k2co3 is produced from reacting 27.9 g ko2 with 29.0 l of c
Airida [17]
The Balanced Chemical Equation is as follow;

                         4 KO₂  +  2 CO₂    →    2 K₂CO₃  +  3 O₂

First find out the Limiting Reagent,
According to equation,

         284 g (4 moles) KO₂ reacted with  =  44.8 L (2 moles) of CO₂
So,
                  27.9 g of KO₂ will react with  =  X  L of CO₂


Solving for X,
                          X  =  (44.8 L × 27.9 g) ÷ 284 g

                          X  =  4.40 L of CO₂

Hence, to consume 27.9 g of KO₂ only 4.40 L CO₂ is required, while, we are provided with 29 L of CO₂, it means CO₂ is in excess and KO₂ is is limited amount, Therefore, KO₂ will control the yield of K₂CO₃. So,

According to eq.

         284 g (4 moles) KO₂ formed  =  138.2 g of K₂CO₃
So,
         27.9 g of KO₂ will form  =  X g of K₂CO₃

Solving for X,
                        X  =  (138.2 g × 27.9 g) ÷ 284 g

                        X  =  13.57 g of K₂CO₃

So, 13.57 g of K₂CO₃ formed is the theoretical yield.

%age Yield  =  13.57 / 21.8 × 100

%age Yield  =  62.24 %
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3 years ago
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Match each of the following topical dosage forms with its correct definition.
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1 year ago
A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci
Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}

The ICE table is then shown as:

                               C_3H_6ClCO_2H_{(aq)}  \ \ \ \ \to  \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ +  \ \ \ \ H^+_{(aq)}

Initial   (M)                     1.8                                       0                               0

Change  (M)                   - x                                     + x                           + x

Equilibrium   (M)            (1.8 -x)                                  x                              x

K_a  = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}

where ;

K_a = 3.02*10^{-5}

3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}

Since the value for K_a is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

3.02*10^{-5} *(1.8) = {(x)(x)}

5.436*10^{-5}= {(x^2)

x = \sqrt{5.436*10^{-5}}

x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M

Dissociated form of  4-chlorobutanoic acid = C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M

Percentage dissociated = \frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100

Percentage dissociated = \frac{7.3729*10^{-3}}{1.8 }*100

Percentage dissociated = 0.4096

Percentage dissociated = 0.41%     (to two significant digits)

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3 years ago
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