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Vika [28.1K]
3 years ago
11

How many moles of Cu(OH)2 are soluble in 1L of sodium hydroxide (NaOH) when the pH is 8.23?

Chemistry
1 answer:
Morgarella [4.7K]3 years ago
7 0

Answer:

4.96E-8 moles of Cu(OH)2

Explanation:

Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.

Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.

pH= -log[H]\\pH= -log (\frac{kw}{[OH]})

8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}

[OH]=1.69E-6

This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:

Cu(OH)_2 -> Cu^{2+} + 2OH^-

In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":

The expression for Kps is:

Kps= [Cu^{2+}] [OH]^2

The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.

Kps= s*(2s+1.69E-6)^2

"s" is the soluble quantity of Cu(OH)2.

The solution for this third grade equation is s=4.96E-8 mol/L

Now, let us calculate the moles in 1 L:

moles Cu(OH)_2 = 4.96E-8 mol/L * 1 L = 4.96E-8 moles

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Perform the calculation and record the answer with the correct number of significant figures.
kotykmax [81]

The question is incomplete, here is the complete question:

A. (6.5-6.10)/3.19

B. (34.123 + 9.60) / (98.7654 - 9.249)

<u>Answer:</u>

<u>For A:</u> The answer becomes 0.1

<u>For B:</u> The answer becomes 0.4884

<u>Explanation:</u>

Significant figures are defined as the figures present in a number that expresses the magnitude of a quantity to a specific degree of accuracy.

Rules for the identification of significant figures:

  • Digits from 1 to 9 are always significant and have infinite number of significant figures.
  • All non-zero numbers are always significant. For example: 664, 6.64 and 66.4 all have three significant figures.
  • All zeros between the integers are always significant. For example: 5018, 5.018 and 50.18 all have four significant figures.
  • All zeros preceding the first integers are never significant. For example: 0.00058 has two significant figures.
  • All zeros after the decimal point are always significant. For example: 2.500, 25.00 and 250.0 all have four significant figures.
  • All zeroes used solely for spacing the decimal point are not significant. For example: 10000 has one significant figure.

<u>Rule applied for addition and subtraction:</u>

The least precise number present after the decimal point determines the number of significant figures in the answer.

<u>Rule applied for multiplication and division:</u>

In case of multiplication and division, the number of significant digits is taken from the value which has least precise significant digits

  • <u>For A:</u> (6.5-6.10)/3.19

This a a problem of subtraction and division.

First, the subtraction is carried out.

\Rightarow \frac{6.5-6.10}{3.19}=\frac{0.4}{3.19}

Here, the least precise number after decimal was 1.

\Rightarrow \frac{0.4}{3.19}=0.125

Here, the least precise number of significant digit is 1. So, the answer becomes 0.1

  • <u>For B:</u> (34.123 + 9.60) / (98.7654 - 9.249)

This a a problem of subtraction, addition and division.

First, the subtraction and addition is carried out.

\Rightarow \frac{34.123+9.60}{98.7654-9.249}=\frac{43.723}{89.5164}=\frac{43.72}{89.516}

Here, the least precise number after decimal in addition are 2 and in subtraction are 3

\Rightarrow \frac{43.72}{89.516}=0.48840

Here, the least precise number of significant digit are 4. So, the answer becomes 0.4884

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