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mafiozo [28]
3 years ago
14

What is the pH of a solution that has a concentration of 1 X 10-3 hydrogen ions?

Chemistry
1 answer:
kykrilka [37]3 years ago
8 0

Given :

Concentration of hydrogen ions, H^+ = 10^{-3} \ ions.

To Find :

The pH of a solution.

Solution :

We know, pH of a solution is given by :

pH = - log [ H^+]\\\\pH = -log ( 10^{-3})\\\\pH = 3

Therefore, the pH of the solution is 3 .

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An atom X has 2 electrons in its outer orbit. What will it do to form a stable ion?
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Answer:

In a neutral molecule, the sum of the bonding valance electrons must be equal. So the products of the negative element and its charges and the positive element and its charge must be equal.

Explanation:

C1×N1 = C2×N2

If we have a 3 valance electrons , the 'A' charge will be either +3 or -5 for a full octet and valance electron in 'B' atoms will mostly result in acquisition of additional electrons (2) for an octet and relative charge of -2.

Balancing the two,

3 × A = -2 × B

To be equal, A = 2 and B = 3

Therefore, A²B³

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History of Aluminum​
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Answer:

Aluminium was named after alum, which is called 'alumen' in Latin. This name was given by Humphry Davy, an English chemist, who, in 1808, discovered that aluminium could be produced by electrolytic reduction from alumina (aluminium oxide), but did not manage to prove the theory in practice.

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How many atoms are in a sample of 1.83 moles of potassium (K) atoms? Please explain the conversions to me. Thank you!
IRINA_888 [86]
1 mole K ------------- 6.02x10²³ atoms
1.83 moles K ------ ?? atoms

1.83 x (6.02x10²³) / 1 =

1.101x10²⁴ atoms of K

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Contributing to greenhouse gases
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The pKa of chloroacetic acid is 2.9. If an organic chemist has a 0.10 M aqueous solution of chloroacetic acid at pH 2.4, what pe
Trava [24]

Answer:

The answer is given below.

Explanation:

We will consider the acid as HA and will set up an ICE table with the equilibrium dissociation of α.

AT pH 2.4 the initial H+ concentration will be 3.98^10-3 M

                      HA →              H+         +      A-

Initial concentration:  0.1    →  3.98 ^10-3       +      0

equilibrium concentration:  0.1(1-α) →   3.98 * 10-3 + 0.1α              0.1α

pKa of chloroacetic acid is 2.9

-log(Ka) = 2.9

Ka = 1.26 * 10-3

From the equation, Ka = [H+] * [A-] / [HA]

1.26 * 10-3 = (3.98 * 10-3 + 0.1α )* 0.1α / 0.1(1-α)      

Since α<<1, we assume 1-α = 1

Solving the equation, we have: α = 0.094

Since this is the fraction of acid that has dissociated, we can say that % of base form = 100 * α= 9.4%

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