Need help with chemistry problem

What is the hydronium ion concentration of a 0.100 M acetic acid solution with Ka = 1.8 × 10-5? The equation for the dissociatio

evablogger [386]

Answer:

1.3×10⁻³ M

Explanation:

Hello,

In this case, given the dissociation reaction of acetic acid:

$CH_3CO_2H(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + CH_3CO_2^-(aq)$

We can write the law of mass action for it:

$Ka=\frac{[H_3O^+][CH_3CO_2^-]}{[CH_3CO_2H]}$

Of course, excluding the water as heterogeneous substances are not included. Then, in terms of the change $x$ due to the dissociation extent, we are able to rewrite it as shown below:

$1.8x10^{-5}=\frac{x*x}{0.100-x}$

Thus, via the quadratic equation or solve, we obtain the following solutions:

$x_1=-0.00135M\\x_2=0.00133M$

Obviously, the solution is 0.00133M which match with the hydronium concentration, thus, answer is: 1.3×10⁻³ M in scientific notation.

Regards.

8 0

4 years ago

Read 2 more answers

Each step in the process below has 60.0% yield

givi [52]

, THR CC14 formed in the first step is used as the reactant used in the second step.if 5.00 mol of CH4 reacts, what is the total amount of HCI producded. assume that C12 an HR in the presentin excess

7 0

3 years ago

Read 2 more answers

8. What do you believe: should husbands make decisions for their

Sedaia [141]

Answer:

I personally don't think that they should because women are perfectly capable of making our own decisions.

Explanation:

hope this helps

7 0

3 years ago

Please, What is the mass of oxygen in 72.0g of water? [ H= 1.0; O = 16.0]

yarga [219]

Convert 72g of water into moles of water using molecular weights.

So water is H2O so add up those molecular weights (H=1 and O=16)

2(1)+(16) = 18 g/mol

Then convert so 72g / (18 g/mol) = 4 mol

Now you can convert mol of water to mol of oxygen. So 4 mol of water is 4 mol of oxygen. Then use oxygen molecular weight to find grams again.

4 mol oxygen * 16 g/mol = 64g of oxygen

If we were doing hydrogen instead of oxygen there would be 8 mol hydrogen in 4 mol of water (2 H’s in every H2O molecule) and since we have 74 grams and oxygen is 64 grams, Hydrogen should be 8 grams. Math to check below

8 mol hydrogen * 1 g/mol = 8g of hydrogen

It all adds to 72 so we are correct.

5 0

3 years ago

The following sequence of reactions occurs in the commercial production of aqueous nitric acid: 4NH3(g)+5O2(g)⟶4NO(g)+6H2O(l)ΔH=

castortr0y [4]

Answer:

The total energy change for the production of one mole of aqueous nitric acid is −494 kJ

Explanation:

Step 1: Data given

4NH3(g)+5O2(g)⟶4NO(g)+6H2O(l)ΔH=−907kJ 2NO(g)+O2(g)⟶2NO2(g)ΔH=−113kJ 3NO2+H2O(l)⟶2HNO3(aq)+NO(g)ΔH=−139kJ

Step 2: Multiply equations

Multiply the first equation by 3:

12 NH3(g) + 15 O2(g) → 12 NO(g) + 18 H2O(l) ΔH = −2721 kJ

Multiply the second equation by 6:

12 NO(g) + 6 O2(g) → 12 NO2(g) ΔH = −678 kJ

Multiply the third equation by 4:

12 NO2(g) + 4 H2O(l) → 8 HNO3(aq) + 4 NO(g) ΔH = −556 kJ

Step 3: Get the equations together

12 NH3(g) + 15 O2(g) + 12 NO(g) + 6 O2(g) + 12 NO2(g) + 4 H2O(l) →

12 NO(g) + 18 H2O(l) + 12 NO2(g) + 8 HNO3(aq) + 4 NO(g)

ΔH = −2721 kJ − 678 kJ − 556 kJ

We can simplify as followed:

12 NH3(g) + 21 O2(g) → 14 H2O(l) + 8 HNO3(aq) + 4 NO(g) ΔH = −3955 kJ

Step 4: Determine the total energy change for the production of one mole of aqueous nitric acid by this process:

−3955 kJ/8 moles HNO3= −494 kJ

The total energy change for the production of one mole of aqueous nitric acid is −494 kJ

3 0

3 years ago