CHEMISTRY COLLEGE

A buffer solution is composed of 4.00 4.00 mol of acid and 3.25 3.25 mol of the conjugate base. If the p K a pKa of the acid is 4.70 4.70 , what is the pH of the buffer?

Answers

Answer 1

Answer:

Answer: The pH of the buffer is 4.61

Explanation:

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

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

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of weak acid = 4.70

$[\text{conjuagate base}]}$ = moles of conjugate base = 3.25 moles

$[\text{acid}]$ = Moles of acid = 4.00 moles

pH = ?

Putting values in above equation, we get:

$pH=4.70+\log((3.25)/(4.00))\n\npH=4.61$

Hence, the pH of the buffer is 4.61

Related Questions

Compare and contrast how observations and results can be used tosupport a conclusion to an experiment.
Draw the predominant product(s) of the following reactions including stereochemistry when it is appropriate. CH3CH2 C C CH3 H2O/H2SO4/HgSO4
(II) To make a secure fit, rivets that are larger than the rivet hole are often used and the rivet is cooled (usually in dry ice) before it is placed in the hole. A steel rivet 1.872cm in diameter is to be placed in a hole 1.870cm in diameter in a metal at 22°C. To what temperature must the rivet be cooled if it is to fit in the hole
When a chemical reaction occurs blank happens
Which is not the name of a family on the periodic tablea) Halogensb) Noble Gasesc) Alkali Earth Metalsd) Actinides

For this reaction, C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O, the ∆H is –2200 kJ. If two moles of C3H8(g) reacted with excess oxygen, what would be true?

Answers

Final answer:

In a reaction where two moles of C3H8 react with an excess of oxygen, the ∆H would be -4400 kJ, double the ∆H value when only one mole is reacted.

Explanation:

The given reaction, C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O, has a ∆H of –2200 kJ. This represents the energy change for the reaction involving one mole of C3H8. When two moles of C3H8 react, the ∆H will be double that of the given ∆H. Therefore, when two moles of C3H8 react with excess oxygen, the ∆H would be -4400 kJ. This is because the ∆H for a reaction is directly proportional to the amount of substance reacted, hence when two moles of C3H8 are reacted, the ∆H is doubled.

Learn more about Chemical Reactions here:

brainly.com/question/34137415

#SPJ12

A chemist titrates 90.0 mL of a 0.5870 M acetic acid (HCH, CO) solution with 0.4794M NaOH solution at 25 °C. Calculate the pH at equivalence. The p Kg of acetic acid is 4.76. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added

Answers

Answer:

9.09

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Which yield comes from measurements obtained during a real experiment?A) actual yield
B) theoretical yield
C) percent yield

Answers

Answer:

A) actual yield

Explanation:

Theoretical yield is the amount of product expected based on the stoichiomety.

Percent yield is the actual yield over the theoretical yield.

What data should be plotted to show that experimental concentration data fits a first-order reaction? A) 1/[reactant] vs. time B) [reactant) vs. time C) In[reactant] vs. time D) In(k) vs. 1/1

Answers

The data that should be plotted to show that experimental concentration data fits a first-order reaction is: C. In [reactant] vs. time.

What is a first-order reaction?

A first-order reaction can be defined as a type of chemical reaction in which the reaction rate (rate of reaction) is directly proportional to the concentration of the reacting chemical substance or elements.

Mathematically, the integrated rate law equation for a first-order reaction is given by this formula:

$A=A_o e^(kt)$

Where:

A is the concentration after time (t).

$A_o$ is the initial concentration of the reactant.

t is the time.

k is the rate constant.

Taking the ln of both sides, we have:

$ln(A)=ln(A_o)-kt$

Therefore, the data that should be plotted to show that experimental concentration data fits a first-order reaction is In[reactant] versus time.

Read more on rate constant here: brainly.com/question/24749252

Answer:

C) In[reactant] vs. time

Explanation:

For a first order reaction the integrated rate law equation is:

$A = A_(0)e^(-kt)$

where A(0) = initial concentration of the reactant

A = concentration after time 't'

k = rate constant

Taking ln on both sides gives:

$ln[A] = ln[A]_(0)-kt$

Therefore a plot of ln[A] vs t should give a straight line with a slope = -k

Hence, ln[reactant] vs time should be plotted for a first order reaction.

of all the hydrogen nuclei in the ocean, 0.0156 how much deuterium could be obtained from 1.0 gal of ordinary tap water

Answers

Answer:

Poop Butt.

Explanation: Poop Butt.

What do lactate dehydrogenase, aspartate aminotransfcrase, and creatine kinase all have in common? a. they all are allosteric enzymes b. they are all zymogens c, they are all used to diagnose medical conditions d. they all function at abeornally high temperatures

Answers

Answer:

c. they are all used to diagnose medical conditions

Explanation:

Lactate dehydrogenase, aspartate aminotransfcrase, and creatine kinase all are used to diagnose medical conditions.

Other Questions

All the following are characteristic properties of phosphorus. Which one is a chemical property?(a) When exposed to air, white phos- phorus will burn spontaneously, but red phosphorus will not.(b) Red phosphorus and white phos- phorus are solids.(c) The white form is soluble in liquid carbon disulfide but is insoluble in water.(d) The red form of phosphorus is in- soluble in both water and carbon disulfide.(e) The red form melts at about 600 ◦ C, and the white form melts at44◦ C.

It is possible for gases in the atmosphere to change the behavior of energy on earth. Agree or disagree

When the following redox equation is balanced with smallest whole number coefficients, the coefficient for Sn(OH)3– will be _____. Bi(OH)3(s) + Sn(OH)3–(aq) → Sn(OH)62–(aq) + Bi(s) (basic solution)

The covalent compounds are soluble in:a) All acidsb) All basesc) all solventsd) nonpolar solvents