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2 years ago

To prepare a lab solution a chemist dissolves solid potassium chloride in a given

Chemistry

1 answer:

Neporo4naja [7]2 years ago

5 0

The solution could be described like a saturated and concentrated solution.

<h3>Solutions</h3>

Dissolution of solutions can occur in certain ways. These ways depend on the solvent dilution coefficient and the amount of solute.

If the amount of solute is less than the solvent can dissolve, the solution is unsaturated.

If the amount of solute is exactly the same, the solution is saturated.

If the amount is greater, the solution is supersaturated.

So, since the amount of solute is exactly equal to the amount that the solvent can dissolve, the solution is saturated and concentrated.

Learn more about solutions in: brainly.com/question/202460

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If you mix 50mL of 0.1 M TRIS acid with 60 mL of0.2 M TRIS base, what will be the resulting pH?

Katyanochek1 [597]

Answer: The pH of resulting solution is 8.7

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

For TRIS acid:

Molarity of TRIS acid solution = 0.1 M

Volume of solution = 50 mL

Putting values in above equation, we get:

$0.1M=\frac{\text{Moles of TRIS acid}\times 1000}{50mL}\\\\\text{Moles of TRIS acid}=0.005mol$

For TRIS base:

Molarity of TRIS base solution = 0.2 M

Volume of solution = 60 mL

Putting values in above equation, we get:

$0.2M=\frac{\text{Moles of TRIS base}\times 1000}{60mL}\\\\\text{Moles of TRIS base}=0.012mol$

Volume of solution = 50 + 60 = 110 mL = 0.11 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of TRIS acid = 8.3

$[\text{TRIS acid}]=\frac{0.005}{0.11}$

$[\text{TRIS base}]=\frac{0.012}{0.11}$

pH = ?

Putting values in above equation, we get:

$pH=8.3+\log(\frac{0.012/0.11}{0.005/0.11})\\\\pH=8.7$

Hence, the pH of resulting solution is 8.7

6 0

3 years ago

Which example indicates that a chemical change has occurred? *

andrezito [222]

Answer:

A. When two aqueous solutions are mixed, a precipitate is formed.

Explanation:

The precipitate (a solid substance that falls from the liquid) is the result of a chemical reaction taking place between the liquids.

The other three answer choices are indicative of physical changes (temperature change, phase change, color change).

7 0

3 years ago

Titration of 0.824 g of potassium hydrogen phthalate required 38.314 g of naoh solution to reach the end point detected by pheno

melamori03 [73]

1.062 mol/kg.

Step 1. Write the balanced equation for the neutralization.

MM = 204.22 40.00

KHC8H4O4 + NaOH → KNaC8H4O4 + H2O

Step 2. Calculate the moles of potassium hydrogen phthalate (KHP)

Moles of KHP = 824 mg KHP × (1 mmol KHP/204.22 mg KHP)

= 4.035 mmol KHP

Step 3. Calculate the moles of NaOH

Moles of NaOH = 4.035 mmol KHP × (1 mmol NaOH/(1 mmol KHP)

= 4.035 mmol NaOH

Step 4. Calculate the mass of the NaOH

Mass of NaOH = 4.035 mmol NaOH × (40.00 mg NaOH/1 mmol NaOH)

= 161 mg NaOH

Step 5. Calculate the mass of the water

Mass of water = mass of solution – mass of NaOH = 38.134 g - 0.161 g

= 37.973 g

Step 6. Calculate the molal concentration of the NaOH

b = moles of NaOH/kg of water = 0.040 35 mol/0.037 973 kg = 1.062 mol/kg

3 0

3 years ago

The larger the hole in the ozone layer gets,

seraphim [82]

The more UV rays reach the earth

4 0

3 years ago

What is the poH of a 2.6 x 10-6 M H+ solution?

andriy [413]

Answer:

Explanation:

pH and pOH.....

The pH is a way of expressing the hydrogen ion concentration.

pH = -log[H+] ............. where [x] means "the concentration of x in moles per liter."

From pH you can compute pOH since at 25C pH + pOH = 14.00 .......... (but only at 25C)

pH = -log(2.6x10^-6) = 5.585 ..... which should be rounded to two significant digits: pH = 5.59

When taking the log of a number, only the digits to the right of the decimal reflect the precision in the original number. Since 2.6x10^-6 has two significant digits, a pH of 5.59 has two significant digits.

pOH + pH = 14.00

pOH = 14.00 - pH = 14.00 - 5.59 = 8.41 ......... at 25C

We can also use the H+ ion concentration to get the hydroxide ion concentration and from that the pOH.

Kw = [H+][OH-] = 1.00x10^-14 .......... at 25C .... like any Kc, the value changes with temperature

[OH-] = Kw / [H+] = 1.00x10^-14 / 2.6x10^-6 = 3.846x10^-9 .... to a couple of guard digits

pOH = -log[OH-] = -log(3.846x10^-9) = 8.415 ...... round to two significant digits: pOH = 8.42 ..... at 25C

=========

Just for grins, you might want to know how Kw changes with temperature, and how [H+] and [OH-] are related at some other temperatures. The pH is the pH of a neutral solution at various temperatures. For instance at 10C a neutral solution has a pH of 7.27. That's not a basic pH. 7.27 is the pH of a neutral solution, but at a different temperature. In a neutral solution at 10C [H+] = [OH-] = 5.41x10^-8M.

pH and Kw for a neutral solution at different temperatures

T .........pH ......... Kw

0......... 7.47....... 0.114 x 10-14

10....... 7.27....... 0.293 x 10-14

20....... 7.08....... 0.681 x 10-14

25....... 7.00....... 1.008 x 10-14

30....... 6.92....... 1.471 x 10-14

40....... 6.77....... 2.916 x 10-14

50....... 6.63....... 5.476 x 10-14

100..... 6.14....... 51.3 x 10-14

4 0

3 years ago