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3 years ago

Write the formula for the conjugate base for the following acids. Thank you

Chemistry

2 answers:

Llana [10]3 years ago

7 0

Answer:

See the answer and explanation below , please.

Explanation:

A conjugate base is defined as that formed after an acid donates its proton.

For each article, a continuation of the conjugate bases (highlighted in bold), for dissociation in water:

a) HF + H20 --> F- + H30+

b) H20+ H20 --> OH- + H30+

C)H2PO3- + H20--> HPO3 2- + H30+

d) HSO4- + H20 --> SO4 2- + H30+

E) HCL02 + H20 --> CLO02 - + H30+

Snezhnost [94]3 years ago

6 0

Answer:

a. HF = F-

b. H2O = H3O+(hydronium ion)

c. H2po3- =HPO3-2 ion

d. HSO4=SO4 (2-)

e. HCLO2= ClO2

Explanation:

The species that is formed is the acid's conjugate base. A more general definition is that a conjugate base is the base member, X-, of a pair of compounds that transform into each other by gaining or losing a proton. The conjugate base is able to gain or absorb a proton in a chemical reaction.

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2 years ago

In a 1.0x10^-4 M solution of HClO(aq), identify the relative molar amounts of these species:HClO, OH-, H3O+, OCl-, H2O

yarga [219]

HClO is a weak acid, which means the ions do not fully dissociate. The hydrolysis reaction for the hypochlorous acid is:

HClO + H2O ⇄ H3O+ +OCl-

Then the equilibrium constant, Ka, of dilute HClO would be:

$K_{a} = \frac{[ H_{3} O^{+} ][O Cl^{-} ]}{HClO}$

Then we do the ICE table. I is for the initial concentration, C for the change and E for the excess.

HClO + H2O ⇄ H3O+ + OCl-
I 1.0x10^-4 0 0
C -x +x +x
E (1.0x10^-4 - x) x x

Substituting the excess (E) concentration to the Ka equation:

$K_{a} = \frac{[x ][x]}{1.0 \ x \ 10^{-4} - x }$

Simplifying the equation would yield a quadratic equation:

$x^{2} + K_{a}x-(1.0 \ x \ 10^{-4}) K_{a}=0$

The Ka for HClO is an experimental data which was determined to be 2.9 x 10^-8. Substitute this to the equation, determine the roots, then you get the value for x, which is the concentration of H3O+ and ClO-. Just use your calculator feature Shift-Solve.

x = 1.688 x 10^-6 M = [H3O+] = [ClO-]

Then, you can determine the conc of [OH-] through pH.

pH = -log {H3O+] = -log [1.688 x 10^-6] = 5.77
pOH = 14 - pH = 14 - 5.77 = 8.23
pOH = 8.23 = -log [OH-]
[OH-] = 5.89 x 10^-9 M

Also, since HClO is (1.0x10^-4 - x), then it's concentration would be:
[HClO] = 1.0x10^-4 - 1.688 x 10^-6 = 9.83 x10^-5 M

Let's summarize all concentrations:
[HClO] = 9.83 x10^-5 M
[OH-] = 5.89 x 10^-9 M
[H3O+] = [ClO-] = 1.688 x 10^-6 M
Since the solution is dilute, H2O is relatively higher in concentration.

Thus in relative amounts, the order would be

H2O >>> HClO > H3O+ = ClO- > OH-

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Write the formula for the conjugate base for the following acids. Thank you ​

Write the formula for the conjugate base for the following acids. Thank you