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2 years ago

What is the ph of a buffer that is 0. 08 m naa and 0. 1 ha. The ka of ha = 1. 0 x 10^-4

1 answer:

Crank2 years ago

4 0

Henderson-Hasselbalch equation relates the pH with the dissociation constant of the acid. The pH of the buffer solution will be 3.90.

<h3>What is the Henderson-Hasselbalch equation?</h3>

The Henderson-Hasselbalch equation is used to calculate the pH or the concentration of the conjugate base and acid.

The Henderson-Hasselbalch equation can be given as,

pH = pKa + log [A⁻] ÷ [HA]

The dissociation reaction is given as,

HA⁺ + H2O ⇌ H3O⁺ + A⁻

NaA → Na⁺ + A⁻

For this first pKa is calculated as:

pKa = - log (1. 0 x 10⁻⁴)

Substituting the value of pKa in Henderson-Hasselbalch equation pH is determined as:

pH = - log (1. 0 x 10⁻⁴) + log [0.08] ÷ [0.1]

= 4 + (-0.0969)

= 3.90

Therefore, 3.90 is the pH of the solution.

Learn more about Henderson-Hasselbalch here:

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1 charged atom is called a Proton

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3 years ago

Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p

UNO [17]

Answer : The equilibrium constant $K_c$ for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of $Br_2$ .

$\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}$

$\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M$

Now we have to calculate the dissociated concentration of $Br_2$ .

The balanced equilibrium reaction is,

$Br_2(g)\rightleftharpoons 2Br(aq)$

Initial conc. 1.731 M 0

At eqm. conc. (1.731-x) (2x) M

As we are given,

The percent of dissociation of $Br_2$ = $\alpha$ = 1.2 %

So, the dissociate concentration of $Br_2$ = $C\alpha=1.731M\times \frac{1.2}{100}=0.2077M$

The value of x = 0.2077 M

Now we have to calculate the concentration of $Br_2\text{ and }Br$ at equilibrium.

Concentration of $Br_2$ = 1.731 - x = 1.731 - 0.2077 = 1.5233 M

Concentration of $Br$ = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

$K_c=\frac{[Br]^2}{[Br_2]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.4154)^2}{1.5233}=0.1133$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.1133

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3 years ago

La columna de la izquierda corresponde a los tipos de sales y la columna derecha a los tipos de fórmula que presentan. Relaciona

Galina-37 [17]

Answer:

1. Hidracidas a. MX

2 Acidas c. MHXO

3. Oxacidas b. MXO

4. Basicas d. M(OH)X

Explanation:

¡Hola!

En este caso, de acuerdo con el concepto de sal, la cual está generalmente dada por la presencia de al menos un metal y un no metal, es posible encontrar cuatro tipos de estas; hidrácidas, oxácidas, básicas y ácidas, en las que las primeras dos son neutras pero la segunda tiene presencia de oxígeno, la tercera tiene iones hidróxido adicionales y la cuarta iones hidrógeno de más.

Debido a la anterior, es posible relacionar cada pareja de la siguiente manera:

1. Hidracidas a. MX

2 Acidas c. MHXO

3. Oxacidas b. MXO

4. Basicas d. M(OH)XO

En las que M se refiere a un metal, X a un no metal, H a hidrógeno y O a oxígeno.

¡Saludos!

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3 years ago

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Answer:

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