All categories

2 years ago

Calculate the pH of a buffer solution prepared by mixing 75.0 mL of 1.00 M lactic acid and 25.0 mL of 0.50 M sodium lactate.

Chemistry

1 answer:

max2010maxim [7]2 years ago

6 0

From the calculations, the pH of the buffer is 3.1.

<h3>What is the pH of the buffer solution?</h3>

The Henderson-Hasselbach equation comes in handy when we deal with the pH of a buffer solution. From that equation;

pH = pKa + log[(salt/acid]

Amount of the salt = 25/1000 * 0.50 M = 0.0125 moles

Amount of the acid = 75/1000 * 1.00 M = 0.075 moles

Total volume = ( 25 + 75)/1000 = 0.1 L

Molarity of salt = 0.0125 moles/0.1 L = 0.125 M

Molarity of the acid = 0.075 moles/0.1 L = 0.75 M

Given that the pKa of lactic acid is 3.86

pH = 3.86 + log( 0.125/0.75)

pH = 3.1

Learn more about pH:brainly.com/question/5102027

#SPJ1

You might be interested in

What is the independent variable, and the dependent variable in this?

Doss [256]

Answer:

The salt and sugar as well as the baby power and granulated sugar are the independent variables.

The dependent variable is how it reacts (will it dissolve or not)

Explanation:

5 0

3 years ago

Read 2 more answers

When the color changes is it a chemical change or a physical change

aev [14]

physical

It can be chemical but it's usually physical as the molecular structure doesn't change

5 0

2 years ago

PLS HELPPP How many moles of oxygen atoms (not molecules) are present in 6.41×1025 molecules of dinitrogen pentoxide (N2O5)? 1.

olga55 [171]

Answer:

(4) 266 moles

Explanation:

We have Dinitrogen Pentoxide N2O5

6.41*10^25 molecules are given

No of Moles of N2O5 = $\frac{No of Molecules }{Avogadro's Number}$

= $\frac{6.41 * 10^25}{6.022*10^23}$

= 106.5 mol

Now using Unitary Method

2 Mole of Nitrogen pentoxide require 5 mole of Oxygen to form N2O5

1 mole of N = $\frac{5}{2}*O$

In 106.5 mole of N = $\frac{5}{2}*106.5$ = 266.25 mole

So, 6.41*10^25 molecules of N2O5 will require 266.25 mole of Oxygen atoms.

5 0

3 years ago

An aqueous solution 10 g of an optically pure substance diluted to 500ml with water and placed in a polarimeter tube 20 cm long.

Gennadij [26K]

Explanation:

Formula to calculate specific rotation is as follows.

Specific rotation ( $[\alpha]$ ) = $\frac{\alpha}{c} \times l$

where, $\alpha$ = observed rotation

c = concentration in g/ml

l = path length in dm

It is given that,

$\alpha = -6.16^{o}$

c = $\frac{10 g}{500 ml}$ = 0.02 g/ml

l = 20 cm = 2 dm (as 1 dm = 10 cm)

Therefore, calculate the specific rotation as follows.

Specific rotation ( $[\alpha]$ ) = $\frac{\alpha}{c} \times l$

= $\frac{-6.16^{o}}{0.02 g/ml} \times 2 dm$

= $-616^{o}$

Thus, we can conclude that the specific rotation of this compound is $-616^{o}$ .

7 0

3 years ago

Help this is due soon!!!

denis23 [38]

Answer:

not 100% sure but I think it's unbalanced chemical

6 0

4 years ago