Calculate the pH of the solution formed when 45.0 mL of 0.100M NaOH solution is added to 50.0 mL of 0.100M CH3COOH (Ka for acetic acid = 1.8 x10-5 ).

Answers

Answer 1

Answer:

The study of chemicals and bonds is called chemistry. There are two types of elements are there and these rare metals and nonmetals.

The correct answer is 5.59.

What is PH?

pH, historically denoting "potential of hydrogen" is a scale used to specify the acidity or basicity of an aqueous solution.
Acidic solutions are measured to have lower pH values than basic or alkaline solutions

The pH of the solution will be due to excessive acid left and the salt formed. Thus, it will form a buffer solution.

The pH of buffer solution is calculated from Henderson Hassalbalch's equation, which is:

$pH= pka+log(salt)/(acid)$

$pka= -log(ka)$

$pka =-log(1.8*10^(-5)$

The moles of acid are taken as:-

$moles = M*V$

$0.5*50=5$

The moles of the base are taken as:-

$moles = M*V$

$0.1*45=4.5$

moles of acid left is 0.5

Place all the values to the equation:-

$pH=4.74+log(4.5)/(0.5) \n=5.69$

Hence, the correct answer is 5.69.

For more information about the pH, refer to the link:-

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Answer 2

Answer:

Answer:

pH of soltion will be 5.69

Explanation:

The pH of the solution will be due to excessive acid left and the salt formed. Thus, it will form a buffer solution.

The pH of buffer solution is calculated from Henderson Hassalbalch's equation, which is:

$pH=pKa+log(([salt])/([acid]) )$

$pKa=-log(Ka)$

$pKa=-log(1.8X10^(-5))=4.74$

The moles of acid taken :

$moles=molarityXvolume=0.1X50=5mmol$

The moles of base taken:

$moles=molarityXvolume=0.1X45=4.5mmol$

The moles of acid left after reaction :

$5-4.5=05mmol$

The moles of salt formed = 4.5mmol

Putting values in equation

$pH=pKa+log(([salt])/([acid]) )=4.74+log((4.5)/(0.5))=5.69$

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Which of the following Ph levels would indicate the weakest base?15862

Consider the following reaction at 298K.I2 (s) + Pb (s) = 2 I- (aq) + Pb2+ (aq)
Which of the following statements are correct?
Choose all that apply.
ΔGo > 0
The reaction is product-favored.
K < 1
Eocell > 0
n = 2 mol electrons
B-

Answers

Answer:

Eªcell > 0; n = 2

Explanation:

The reaction:

I2 (s) + Pb (s) → 2 I- (aq) + Pb2+ (aq)

Is product favored.

A reaction that is product favored has ΔG < 0 (Spontaneous)

K > 1 (Because concentration of products is >>>> concentration reactants).

Eªcell > 0 Because reaction is spontaneous.

And n = 2 electrons because Pb(s) is oxidizing to Pb2+ and I₂ is reducing to I⁻ (2 electrons). Statements that are true are:

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2. Calculate the mass of solvent in grams in a solution containing 3.0 grams of Tylenolif the mass percent is 3.5%.

Answers

Answer:

83g

Explanation:

1, Formula

Mass percent = (mass of solute/mass of solution) × 100

2. Determine mass of solution

Substitute the data and clear the mass of solution:

3.5 = (3.0g / mass of solution) × 100

0.035 = 3.0g / mass of solution

mass of solution = 3.0 g / 0.035

mass of solution = 85.714g

3. Determine the mass of solvent:

mass of solvent = mass of solution - mass of solute

mass of solvent = 85.714g - 3.0g = 82.714g

Round to two significant figures: 83 g

Rank the following compounds in order of decreasing acid strength using periodic trends. Rank the acids from strongest to weakest. To rank items as equivalent, overlap them. H2Se

HBr

H2O

HI

Answers

Explanation:

It is known that acidic strength of hydrides of same group tends to increase when we move from top to bottom in a group. On the other hand, acidic strength of hydrides of same period elements increases when we move from left to right in a period.

As both bromine and iodine belongs to the same group. Also, selenium and oxygen are same group elements. Therefore, their acidic strength increases on moving down the group.

Therefore, we can conclude that acidic strength of given compounds from strongest to weakest is as follows.

HI > HBr > $H_(2)Se$ > $H_(2)O$

Final answer:

To rank the acids in decreasing acid strength using periodic trends, consider the size, electronegativity, and presence of lone pairs of electrons. HI is the strongest acid, followed by HBr, H2O, and H2Se.

Explanation:

To rank the acids in order of decreasing acid strength using periodic trends, we need to consider the size and electronegativity of the atoms. The larger the atom, the weaker the acid, and the more electronegative the atom, the stronger the acid. Additionally, we can consider the presence of lone pairs of electrons, as they increase the acidity.

HI - Iodine (I) is larger and less electronegative than the other halogens. It also has a lone pair of electrons, making it the strongest acid.
HBr - Bromine (Br) is larger and less electronegative than chlorine (Cl), and it also has a lone pair of electrons. It is the second strongest acid.
H2O - Oxygen (O) is smaller and more electronegative than the halogens. It does not have a lone pair of electrons, making it a weaker acid than the halogens.
H2Se - Selenium (Se) is larger and less electronegative than sulfur (S). However, it does not have a lone pair of electrons, making it the weakest acid.

Learn more about Periodic trends here:

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What is the question mark

Answers

It is likely asking you to determine the coefficient on the chemical and balance the equation.

Which one of the following compounds will NOT be soluble in water? Which one of the following compounds will NOT be soluble in water? LiOH BaSO4 NaNO3 MgCl2 K2S

Answers

Answer:

$BaSO_(4)$ will be not soluble in water

Explanation:

LiOH is a strong base. Hence it gets completely dissociated in aqueous solution.

$NaNO_(3)$ is a strong electrolyte. Hence it gets completely dissociated in aqueous solution.

$MgCl_(2)$ is a strong electrolyte. Hence it gets completely dissociated in aqueous solution.

$K_(2)S$ is a strong electrolyte. Hence it gets completely dissociated in aqueous solution.

$BaSO_(4)$ is a sparingly soluble salt. Hence it is not dissociated and hence dissolved in water. This is due to the fact that both $Ba^(2+)$ and $SO_(4)^(2-)$ ions are similar in size. Hence crystal structure of $BaSO_(4)$ is quite stable. Hence $BaSO_(4)$ is reluctant to undergo any dissociation in aqueous solution.

What is the solubility in pure water of ba(io3)2 in moles per liter at 25 ˚c? [ksp (25 ˚c) = 6.0  10–10]?

Answers

Ba(IO₃)₂(s) partially dissociates in water as Ba²⁺(aq) and IO₃⁻(aq).
Ba(IO₃)₂(s) ⇄ Ba²⁺(aq) + 2IO₃⁻(aq)
Initial
Change -X +X 2X
Equilibrium X 2X

Ksp = [Ba²⁺(aq)] x [IO₃⁻(aq)]²
6.0 x 10⁻¹⁰ = X * (2X)²
6.0 x 10⁻¹⁰ = 4X³
X = 5.313 x 10⁻⁴ mol/L

Hence, the solubility of the Ba(IO₃)₂(s) is 5.313 x 10⁻⁴ mol/L

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