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mezya [45]
3 years ago
8

A 1.0L buffer solution contains 0.100 molHC2H3O2 and 0.100 molNaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5. Calculate the p

H of the solution upon the addition of 10.0 mL of 1.00 MHCl to the original buffer.
Express the your answer to two decimal places.
***will mark Brainliest for right answer ***
Chemistry
1 answer:
Crazy boy [7]3 years ago
7 0
Starting pH = pKa because the acid and conjugate base have equal concentrations.
pH = pKa = -log(1.8*10–5) = 4.74

0.0100 L • 1.00 M = 0.0100 mol HCl, which reacts with the conjugate base, producing 0.0100 mol of weak acid.

You now have 0.110 mol weak acid and 0.090 mol conjugate base. Use the Henderson-Hasselbach equation to solve for pH now.

pH = 4.74 + log(0.090/0.110) = 4.65

The equation technically says to use concentrations of the acid and base, but that doesn’t actually matter, because it would give the same answer.
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Look at the photo.

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6 0
2 years ago
For the reaction: 3 H2(g) + N2(g) <--> 2 NH3(g), the concentrations at equilbrium were [H2] = 0.10 M, [N2] = 0.10 M, and [
masya89 [10]

The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵

<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>

It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.

Hence, the equilibrium constant of the reaction in discuss is;

K = [5.6]²/[0.10]³[0.10]

k = 5.6² × 10⁴

k = 3.136 × 10⁵

K = 3.1 × 10⁵.

Read more on equilibrium constant;

brainly.com/question/1619133

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5 0
2 years ago
How many liters of 3.5 M solution can be made using 23 moles of LiBr *must show work to get credit*
MrMuchimi

<u>Answer:</u> 6.57 L of solution can be made.

<u>Explanation:</u>

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

\text{Molarity of solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (L)}} .....(1)

Given values:

Molarity of LiBr = 3.5 M

Moles of LiBr = 23 moles

Putting values in equation 1, we get:

3.5mol/L=\frac{23mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{23mol}{3.5mol/L}=6.57L

Hence, 6.57 L of solution can be made.

5 0
3 years ago
37.5g of nitrogen reacts with 15.5 g of hydrogen. What mass of ammonia can be made? What is the limiting reactant?
Gennadij [26K]

Answer:

Limiting reactants or limiting reagents decide the amount of product formed and the amount of excess reagent used.

7 0
3 years ago
Iron (III) bromide reacts with sodium hydroxide to produce iron (III) hydroxide and sodium bromide. Write the balanced chemical
valkas [14]
3 NaOH + FeBr3 = 3 NaBr + Fe(OH)3

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3 0
3 years ago
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