All categories

3 years ago

Explain why a strong acid must have a weak conjugate base

Chemistry

1 answer:

mojhsa [17]3 years ago

4 0

Explanation:

The specie produced when a strong acid is dissociated in water along with the proton is the conjugate base of that acid.

Consider a strong acid such as HCl which donates the proton so readily as it is a strong base and there is no tendency for conjugate base, which is $Cl^-$ to re accept that proton.

A strong base is the one which accepts proton and holds it firmly. Thus, the strong acid has weak conjugate base.

You might be interested in

A solution at 25 degrees Celslus 1.0*10^ -5 MH 3 O^ + . What the concentration of OH^ - in this solution?

dexar [7]

Answer:

Explanation:

-log(1.0x10^-5) = pH

pH + pOH = 14 (rearrange it)

OH- = 10^-pOH = 1.0 x 10^-9

- Hope that helped! Let me know if you need further explantion.

7 0

3 years ago

Mass does not change during a chemical change. is the same true for physical change?

Mashutka [201]

Answer:

Mass states that mass is neither created nor destroyed in a chemical reaction or a physical transformation

Explanation:

4 0

3 years ago

For 100.0 mL of a solution that is 0.040M CH3COOH and 0.010 M CH3COO, what would be the pH after adding 10.0 mL 50.0 mM HCl?

damaskus [11]

Answer:

The pH of the buffer is 3.90

Explanation:

The mixture of a weak acid CH3COOH and its conjugate base CH3COO produce a buffer that follows the equation:

pH = pKa + log [A-] / [HA]

Where pH is the pH of the buffer, pKa is the pKa of acetic acid (4.75), and [A-] could be taken as the moles of the conjugate base and [HA] the moles of thw weak acid.

To solve this question we need to find the moles of the CH3COOH and CH3COO- after the reaction with HCl:

CH3COO- + HCl → CH3COOH + Cl-

The moles of CH3COO- are its initial moles - the moles of HCl added

And moles of CH3COOH are its initial moles + moles HCl added

Moles CH3COO-:

Initial moles = 0.100L * (0.010mol / L) = 0.00100moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.000500 moles

Moles CH3COOH:

Initial moles = 0.100L * (0.040mol / L) = 0.00400moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.003500 moles

pH is:

pH = 4.75 + log [0.000500] / [0.00350]

pH = 3.90

<h3>The pH of the buffer is 3.90</h3>

3 0

3 years ago

How many grams of ammonia must you start with to make 900.00 l of a 0.140 m aqueous solution of nitric acid? assume all the reac

bogdanovich [222]

You need the set of reactions that goes from ammonia to nitric acid.

1) 4NH3(g)+5O2(g)-->4NO(g)+6H2O(g)

2) 2NO(g)+O2(g)-->2NO2(g)

3) 3NO2(g)+H2O(l)-->2HNO3(aq)+NO(g)

State the ratio of moles of HNO3 to NH3:

4 moles of NH3 produce 4 mole of NO,

4 moles of NO produce 4 moles of NO2

4 moles of NO2 produce 4 * (2 / 3) moles of HNO3 = 8/3 moles of HNO3.

=> (8/3) moles HNO3 : 4 moles NH3

Calculate the number of moles of HNO3 in 900.00 l of 0.140 M solution

M = n / V => n = M * V = 0.140 M * 900.00 liter = 126 moles HNO3

Use proportions:

(8/3) moles HNO3 / 4 moles NH3 = 126 moles HNO3 / x

=> x = 126 moles HNO3 * 4 moles NH3 / (8/3 moles HNO3) = 189 moles NH3

Convert moles to grams:

molar mass NH3 = 14 g/mol + 3 * 1g/mol = 17 g/mol

mass in grams = number of moles * molar mass = 189 moles * 17 g/mol = 3213 g

Answer: 3213 g.

3 0

4 years ago

Which of the drawings contain a mixture?

Assoli18 [71]

the answer is c

Explanation:

because c has all elements mixed together

4 0

3 years ago