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guajiro [1.7K]
2 years ago
8

You have a 250. -ml sample of 1. 28 m acetic acid (ka = 1. 8 x´ 10–5). calculate the ph of the best buffer.

Chemistry
2 answers:
exis [7]2 years ago
7 0

The ideal buffer's pH value is 4.74..

What is best buffer ?

The Best Buffer is a buffer that has equal amounts of the weak acid and its conjugate base. It responds well to the addition of acids and bases in order to fend off pH changes. A mixture of the weak base and its conjugate acid, or the weak acid and its conjugate base, is referred to as a buffer solution.

Here weak acid is acetic acid CH3COOH

The following equation can be used to determine a buffer's pH:

pH = pka + logconjugate baseweak acid

Now, the weak acid and its conjugate base are present in equal amounts in the optimal buffer.

Therefore,

Conjugate base = weak acid

This implies that,

pH = pka + log 1

pH = pka  + 0

pH = pka

Now, pka can be written as -log ka

The value of ka is given as 1.8 * 10-5

Therefore,

pH = -log (1.8 * 10-5)

pH = 4.74

As a result, the ideal buffer has a pH of 4.74..

Learn more about the pH on

brainly.com/question/22390063

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Nadya [2.5K]2 years ago
4 0

The ph of the best buffer is 4.74

The given acetic acid is a weak acid

The equation of the pH of the buffer

pH = pKa + log ( conjugate base / weak acid ).

For best buffer the concentration of the weak acid and its conjugate base is equal.

pH = pKa + log 1

pH = pKa + 0

pH = pKa

given Ka = 1.8 × 10⁻⁵

pKa = - log ka

pH = -log ( 1.8 × 10⁻⁵ )

pH = 4. 74

Hence the pH of the best buffer is 4.74

Learn more about the pH on

brainly.com/question/22390063

#SPJ4

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<u>Answer:</u> The molarity of barium hydroxide solution is 0.118 M.

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HClO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Ba(OH)_2

We are given:

n_1=1\\M_1=0.102M\\V_1=24.0mL\\n_2=2\\M_2=?M\\V_2=10.3mL

Putting values in above equation, we get:

1\times 0.102\times 24.0=2\times M_2\times 10.3\\\\M_2=0.118M

Hence, the molarity of Ba(OH)_2 solution will be 0.118 M.

6 0
3 years ago
What volume of 12 M NaOH and 2 M NaOH should be mixed to get 2 litres of 9 M NaOH solution?
Sergio [31]
8/5lit.. of 12M NaOH
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3 years ago
If the reaction N2 (g) + 3 H2 (g) --&gt; 2 NH3 (g) has the concentrations 1.1 M for nitrogen, 0.75 M for hydrogen and 0.25 M fo
Luba_88 [7]

<u>Answer:</u> The value of K_c is 0.136 and is reactant favored.

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{c}

For the chemical reaction between carbon monoxide and hydrogen follows the equation:

N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)

The expression for the K_{c} is given as:

K_{c}=\frac{[NH_3]^2}{[N_2][H_2]^3}

We are given:

[NH_3]=0.25M

[H_2]=0.75M

[N_2]=1.1M

Putting values in above equation, we get:

K_c=\frac{(0.25)^2}{1.1\times (0.75)^3}

K_c=0.135

There are 3 conditions:

  • When K_{c}>1; the reaction is product favored.
  • When K_{c}; the reaction is reactant favored.
  • When K_{c}=1; the reaction is in equilibrium.

For the given reaction, the value of K_c is less than 1. Thus, the reaction is reactant favored.

Hence, the value of K_c is 0.136 and is reactant favored.

4 0
3 years ago
The molar heat capacity of an unknown substance is 92.1 J/mol-K. If the unknown has a molar mass of 118 g/mol, what is the speci
dlinn [17]

Answer : The specific heat (J/g-K) of this substance is, 0.780 J/g.K

Explanation :

Molar heat capacity : It is defined as the amount of heat absorbed by one mole of a substance to raise its temperature by one degree Celsius.

1 mole of substance releases heat = 92.1 J/K

As we are given, molar mass of unknown substance is, 118 g/mol that means, the mass of 1 mole of substance is, 118 g.

As, 118 g of substance releases heat = 92.1 J/K

So, 1 g of substance releases heat = \frac{92.1}{118}=0.780J/g.K

Thus, the specific heat (J/g-K) of this substance is, 0.780 J/g.K

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