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boyakko [2]
3 years ago
14

A buffer is composed of nh3 and nh4cl. How would this buffer solution control the ph of a solution when a small amount of a stro

ng base is added?
Chemistry
1 answer:
Shkiper50 [21]3 years ago
5 0

Answer:

According to Le-chatelier principle, equilibrium will shift towards left to minimize concentration of OH^{-} and keep same equilibrium constant

Explanation:

In this buffer following equilibrium exists -

NH_{3}(aq.)+H_{2}O(l)\rightleftharpoons NH_{4}^{+}(aq.)+OH^{-}(aq.)

So, OH^{-} is involved in the above equilibrium.

When a strong base is added to this buffer, then concentration of OH^{-} increases. Hence, according to Le-chatelier principle, above equilibrium will shift towards left to minimize concentration of OH^{-} and keep same equilibrium constant.

Therefore excess amount of OH^{-} combines with NH_{4}^{+} to produce ammonia and water. So, effect of addition of strong base on pH of buffer gets minimized.

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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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