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marta [7]
3 years ago
10

Solution A is a 1.00 L buffer solution that is 1.188 M in acetic acid and 1.188 M in sodium acetate. Acetic acid has a pKa of 4.

74. What is the pH change of this solution upon addition of 0.1 mol of HCl? Enter your answer numerically to four decimal places. It will be a negative number.
Chemistry
1 answer:
Tomtit [17]3 years ago
7 0

Answer:

pH change is -0.07

Explanation:

Using H-H equation for acetic acid:

pH = pKa + log [Acetate salt] / [Acetic acid]

Replacing:

pH = 4.74 + log[1.188M] / [1.188M]

pH = 4.74

The HCl reacts with sodium acetate producing acetic acid, thus:

HCl + CH₃COONa → CH₃COOH + NaCl

That means the final moles of sodium acetate are initial moles - moles of HCl and moles of acetic acid are initial moles + moles of HCl.

As the volume of the buffer is 1.0L, initial moles of both substances are 1.188moles. After reaction, the moles are:

sodium acetate: 1.188mol - 0.1mol = 1.088mol

Acetic acid: 1.188mol + 0.1mol = 1.288mol

Using again H-H equation:

pH = 4.74 + log[1.088M] / [1.288M]

pH = 4.67

pH change is: 4.67 - 4.74 = -0.07

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Lera25 [3.4K]

Answer:

NH3

Explanation:

2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l)

So for two moles of NH3 we need one mole of CO2. So let's count moles for each reagent.

n(NH3)=m(NH3)/M(NH3)=135700/17,03=7968.29 mol

n(CO2)=m(CO2)/M(CO2)=211400/44.01=4803.45 mol

From equation we have to divide n(NH3) by 2 because we need two equivalent per one CO2. That will be 3984.145. So the limiting agent is NH3 because it's not enough of it to react with all CO2

4 0
3 years ago
What is the ph of a solution labeled .30 trimethylamine k for trimethylamine is 7.42 x 10^4
valentina_108 [34]

Answer:the pH is 12

Explanation:

First We need to understand the structure of trimethylamine

Due to the grades of the bond in the nitrogen with a hybridization sp3 is 108° approximately, then is generated a dipole magnetic at the upper side of the nitrogen, this dipole magnetic going to attract a hydrogen molecule of the water making the water more alkaline

C3H9N+ H2O --> C3H9NH + OH-

k=\frac{[C3H9NH]*[OH-]}{[C3H9N]}

Then:

The concentration of the trimethylamine is 0.3 and the concentration of the ion C3H9NH is equal to the OH- relying on the stoichiometric equation. We could find the concentration of the OH- ion with the square root of the multiplication between k and the concentration of trimethylamine

[OH-]=\sqrt{ 0.3*7.42x10^{-4}}

[OH-]=0.01

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This is because the reactants are typically always on the left side of the yields symbol. In this decomposition reaction, it would still be the same as at the end of the reaction, there were to products produced: Mercury and Oxygen.

Products tend to always be on the right side of the yields symbol, they're what comes out of a reaction no matter what type.

Hope this helps!

7 0
4 years ago
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By using a conversion factor:

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In order to get the grams of IKI to create a 100 mL solution of 0.600 M IKI, use the same formula as above:

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