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2 years ago

What is the ph of a buffer that is 0. 08 m naa and 0. 1 ha. The ka of ha = 1. 0 x 10^-4

1 answer:

Crank2 years ago

4 0

Henderson-Hasselbalch equation relates the pH with the dissociation constant of the acid. The pH of the buffer solution will be 3.90.

<h3>What is the Henderson-Hasselbalch equation?</h3>

The Henderson-Hasselbalch equation is used to calculate the pH or the concentration of the conjugate base and acid.

The Henderson-Hasselbalch equation can be given as,

pH = pKa + log [A⁻] ÷ [HA]

The dissociation reaction is given as,

HA⁺ + H2O ⇌ H3O⁺ + A⁻

NaA → Na⁺ + A⁻

For this first pKa is calculated as:

pKa = - log (1. 0 x 10⁻⁴)

Substituting the value of pKa in Henderson-Hasselbalch equation pH is determined as:

pH = - log (1. 0 x 10⁻⁴) + log [0.08] ÷ [0.1]

= 4 + (-0.0969)

= 3.90

Therefore, 3.90 is the pH of the solution.

Learn more about Henderson-Hasselbalch here:

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The principal component of mothballs is naphthalene, a compound with a molecular mass of about 130 amu, containing only carbon a

DIA [1.3K]

Answer:

Empirical formula = C5H4

Molecular formula = C10H8

Explanation:

When the 3000 mg of naphthalene are burned they produce 10.3 mg of CO2. Knowing the unbalanced equation of the combustion of naphthalene, we have:

CxHy + O2 = CO2 + H2O

We calculate the molar composition of the sample. We look for the molecular weights in the periodic table:

CO2 = 12,011 + 2 (15,999) = 44,009 g

Mol C = 10.3 mg * (1 mol CO2 / 44.009 g CO2) * (1 mol C / 1 mol CO2) = 0.234 mmol C

Mass C = 0.234 mmol C * (12.011 g C / 1 mol C) = 2.8105 mg C

Mass H = 3 mg - 2.8105 mg = 0.1895 mg H

Mol H = 0.1895 mg H * (1 mol H / 1,008 g H) = 0.188 mmol H

To calculate the empirical formula, we must divide the number of moles of each element by the smallest number of moles, in this case, of hydrogen:

C = 0.2340 mmol C / 0.1895 mol H = 1.25

H = 0.1895 mmol H / 0.1895 mmol H = 1

We multiply the coefficients by 4, and we have the empirical formula:

C1.25 * 4H1 * 4 = C5H4

The molecular formula is equal to (C5H4)m, where m is calculated by the molecular and empirical mass ratio, as follows:

Empirical mass = (5 * 12.011) + (4 * 1.008) = 64.09 g

m = 130 g / 64.09 g = 2.02 = 2

Therefore we have the molecular formula:

(C5H4)2 = C10H8

4 0

3 years ago

When we combine two simpler substances into a more complex one, it is known as a reaction.

Nata [24]

Answer:

It would be known as a chemical reaction or a chemical compound.

3 0

3 years ago

A chemical ____ is a process in which some substances change chemically into different substances

garik1379 [7]

Is reaction a choice?

6 0

2 years ago

Read 2 more answers

Using the redox reaction below determine which element is oxidized and which is reduced. 4nh3 + 3ca(clo)2 → 2n2 + 6h2o + 3cacl2

lys-0071 [83]

The oxidation number of elements in equation below are,

4NH₃ + 3Ca(ClO)₂ → 2N₂ + 6H₂O + 3CaCl₂

O.N of N in NH₃ = -3
O.N of Ca in Ca(ClO)₂ and CaCl₂ = +2
O.N of N in N₂ = 0
O.N of Cl in Ca(ClO)₂ = +1
O.N of Cl in CaCl₂ = -1

Oxidation:
Oxidation number of Nitrogen is increasing from -3 (NH₃) to 0 (N₂).

Reduction:
Oxidation number of Cl is decreasing from +1 [Ca(ClO)₂] to -1 (CaCl₂).

Result:
<span>N is oxidized and Cl is reduced.</span>

6 0

3 years ago

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A student obtains a mixture of the chlorides of two unknown metals, X and Z. The percent by mass of X and the percent by mass of

Aleksandr-060686 [28]

<u>Answer:</u> The additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X

<u>Explanation:</u>

To calculate the mole percent of a substance, we use the equation:

$\text{Mole percent of a substance}=\frac{\text{Moles of a substance}}{\text{Total moles}}\times 100$

Mass percent means that the mass of a substance is present in 100 grams of mixture

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

We require the molar masses of Z and X to calculate the mole percent of Z and X respectively

Hence, the additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X

8 0

3 years ago