Using chemical equations, show how the triprotic acid h3po4 ionizes in water. phases, such as (l) or (aq), are optional.

Phosphate acid includes Triprotic acid and weak acid, which will be ionized in 3 stages

According to Arrhenius, acids are substances which, when dissolved in water, release Hions.

An HₓY acid in water will ionize:

HₓY (aq) --------> xH⁺ (aq) + Yˣ- (aq)

Example:

HCl -------> H⁺ + Cl⁻

The amount of Hions produced by 1 acid molecule is called valence acid, whereas acidic residuals are formed after the release of Hions.

Usually, the name acid begins with the word acid followed by the name of the remaining acidic ion

The ion concentration of a weak acid is determined by the value of the acid ionization constant (Ka).

The greater the value of Ka, the greater the dissociated acid produces its Hion and the greater its acidity

Based on the number of protons (H ⁺) released, acids can be divided into:

• 1. Monoprotic acid is an acid which releases one H + ion such as HCl, HCN, HNO₃

• 2. Polyprotic acid which is an acid that releases more than one H⁺ ion.

This acid can be divided into two:

In general, the weak acid ionization reaction

HA (aq) ---> H⁺ (aq) + A⁻ (aq)

Ka's value

Phosphate acid includes Triprotic acid and weak acid which will be ionized in 3 stages  :

Ka = Ka₁ x Ka₂ x Ka₃

What is the reaction to HCN  

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base would not effectively deprotonate acetylene  

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strong enough acids  

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it is important to conserve water because there is very limited amount of freshwater here on earth and if we are not smart about how we use it we will only be runnning out of our supply faster than neccessarry.

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years