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Olenka [21]
3 years ago
14

If a sample of a certain solution is determined to have a [H3O+] concentration of 9.49×10−1 moles/liter, what is its pH? Round o

ff your answer to one decimal place.
Chemistry
1 answer:
almond37 [142]3 years ago
6 0

<u>Answer:</u> The pH of the solution is 1.022

<u>Explanation:</u>

pH is defined as the negative logarithm of hydrogen ion concentration present in a solution.

To calculate the pH of the reaction, we use the equation:

pH=-\log[H^+]

where,

[H^+]=9.49\times 10^{-1}mol/L

Putting values in above equation, we get:

pH=-\log(9.49\times 10^{-1})\\\\pH=1.022

Hence, the pH of the solution is 1.022

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Give the number of protons and neutrons in the nucleus of the following atoms
damaskus [11]

The number of protons and neutrons in the nucleus​ : 92 and 146

<h3>Further explanation</h3>

Given

Mass number = 238

Atomic number = 92

Required

Protons and neutrons

Solution

Mass number=number of protons + neutrons

Atomic number=number of protons=number of electrons

Atomic number = 92⇒number of protons=92

Number of neutrons = mass number - atomic number

Number of neutrons = 238 - 92 = 146

3 0
2 years ago
Phosphoric acid is a triprotic acid with the following pKa values:pKa1=2.148, pKa2=7.198, pKa3=12.375You wish to prepare 1.000 L
laila [671]

Answer:

NaH₂PO₄ =  1.876 g

Na₂HPO₄ =  4.879 g

Combinations: H₃PO₄ and Na₂HPO₄; H₃PO₄ and Na₃HPO₄

Explanation:

To have a buffer at 7.540, the acid must be in it second ionization, because the buffer capacity is pKa ± 1. So, we must use pKa2 = 7.198

The relation bewteen the acid and its conjugated base (ion), is given by the Henderson–Hasselbalch equation:

pH = pKa + log[A⁻]/[HA], where [A⁻] is the concentration of the conjugated base, and [HA] the concentration of the acid. Then:

7.540 = 7.198 + log[A⁻]/[HA]

log[A⁻]/[HA] = 0.342

[A⁻]/[HA] = 10^{0.342}

[A⁻]/[HA] = 2.198

[A⁻] = 2.198*[HA]

The concentration of the acid and it's conjugated base must be equal to the concentration of the buffer 0.0500 M, so:

[A⁻] + [HA] = 0.0500

2.198*[HA] + [HA] = 0.0500

3.198*[HA] = 0.0500

[HA] = 0.01563 M

[A⁻] = 0.0500 - 0.01563

[A⁻] = 0.03436 M

The mix reaction is

NaH₂PO₄ + Na₂HPO₄ → HPO₄⁻² + 3Na + H₂PO₄⁻

The second ionization is:

H₂PO₄⁻ ⇄ HPO₄⁻² + H⁺

So, H₂PO₄⁻ is the acid form, and its concentration is the same as NaH₂PO₄, and HPO₄⁻² is the conjugated base, and its concentration is the same as Na₂HPO₄ (stoichiometry is 1:1 for both).

So, the number of moles of these salts are:

NaH₂PO₄ = 0.01563 M * 1.000 L = 0.01563 mol

Na₂HPO₄ = 0.03436 M* 1.000 L = 0.03436 mol

The molar masses are, Na: 23 g/mol, H: 1 g/mol, P: 31 g/mol, and O = 16 g/mol, so:

NaH₂PO₄ = 23 + 2*1 + 31 + 4*16 = 120 g/mol

Na₂HPO₄ = 2*23 + 1 + 31 + 4*16 = 142 g/mol

The mass is the number of moles multiplied by the molar mass, so:

NaH₂PO₄ = 0.01563 mol * 120 g/mol = 1.876 g

Na₂HPO₄ = 0.03436 mol * 142 g/mol = 4.879 g

To prepare this buffer, it's necessary to have in solution the species H₂PO₄⁻ and HPO₄⁻², so it can be prepared for mixing the combination of:

H₃PO₄ and Na₂HPO₄ (the acid is triprotic so, it will form the H₂PO₄⁻ , and the salt Na₂HPO₄ will dissociate in Na⁺ and HPO₄²⁻);

H₃PO₄ and Na₃HPO₄ (same reason).

The other combinations will not form the species required.

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The hotter it is, the faster the atoms in something are moving.

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