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2 years ago

6

Explain why the dissociation of a weak acid can be ignored when calculating the pH of a solution that contains both a weak acid

and a strong acid. The ion from the strong acid is a common ion in the dissociation reaction of the weak acid. Consequently, it shifts the equilibrium back toward the which the dissociation of the to the point where the amount of ion that it produces is negligible

1 answer:

Sliva [168]2 years ago

4 0

Answer:

The answer is in the explanation.

Explanation:

The dissociation of a weak acid consist in the following equilibrium:

HX ⇄ H⁺ + X⁻

Where Ka is defined as:

Ka = [H⁺] [X⁻] / [HX]

A strong acid (HY) dissociates completely in water, thus:

HY → H⁺ + Y⁻

As the strong acid produces H⁺, in the equilibrium, the reaction shifts to the left -The undissociated form-, reducing the production of H⁺, allowing ignore the dissociation of the weak acid when calculating the pH.

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which Substance can be used to Neutralize nitric acid? A Acetic acid C Hydrogen peroxide b milk of magnesia D Ethanol

sweet-ann [11.9K]

Answer:B

Explanation:

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2 years ago

A chemist prepares a solution of sodium chloride by measuring out 25.4 grams of sodium chloride into a 100. mL volumetric flask

mr_godi [17]

Answer:

The concentration in mol/L = 4.342 mol/L

Explanation:

Given that :

mass of sodium chloride = 25.4 grams

Volume of the volumetric flask = 100 mL

We all know that the molar mass of sodium chloride NaCl = 58.5 g/mol

and number of moles = mass/molar mass

The number of moles of sodium chloride = 25.4 g/58.5 g/mol

The number of moles of sodium chloride = 0.434188 mol

The concentration in mol/L = number of mol/ volume of the solution

The concentration in mol/L = 0.434188 mol/ 100 × 10⁻³ L

The concentration in mol/L = 4.34188 mol/L

The concentration in mol/L = 4.342 mol/L

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2 years ago

Elements with atomic numbers 7 and 83 have how many valence electrons? *

IRINA_888 [86]

Explanation:

element with atomic numbers 7 and 83 have 5 and 5 valence electrons

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2 years ago

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

<u>Part (a) :</u>

<u>Given:</u>

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

<u>Part (b) :</u>

<u>Given:</u>

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

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3 years ago

Your tongue and nose work separately.<br> True<br> False

yuradex [85]

False, our tongue and nose work together

3 0

2 years ago

Read 2 more answers