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ziro4ka [17]
3 years ago
14

At a certain temperature, the pHpH of a neutral solution is 7.45. What is the value of KwKwK_w at that temperature? Express your

answer numerically using two significant figures. View Available Hint(s)
Chemistry
2 answers:
Darya [45]3 years ago
7 0

Answer:

The value of Kw is 1.26 * 10^-15

Explanation:

Step 1: Data given

The pH of the neutral solution = 7.45

pH = -log[H+]

Kw = [OH-][H+]

Step 2: Calculate [H+]

pH = 7.45

-log[H+] = 7.45

[H+] =  10^-7.45

[H+] = 3.55 * 10^-8 M

Step 3: Calculate [OH-]

Since pH and pOH are equal in our neutral solution so are [H+] = [OH-]

[OH-] = 3.55 * 10^-8 M

Step 4: Calculate Kw

Kw = [H]*[OH]

Kw = 3.55 * 10^-8 M * 3.55 * 10^-8 M

Kw = (3.55 * 10^-8)²

Kw = 1.26 * 10^-15

The value of Kw is 1.26 * 10^-15

Lerok [7]3 years ago
4 0

Answer:

1.3 × 10⁻¹⁵

Explanation:

At a certain temperature, the pH of a neutral solution is 7.45. Then, we can calculate the concentration of H⁺.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -7.45

[H⁺] = 3.55 × 10⁻⁸

By definition, in a neutral solution the concentration of H⁺ is equal to that of OH⁻.

Finally, we can calculate the ionic product of water (Kw).

Kw = [H⁺] × [OH⁻]

Kw = 3.55 × 10⁻⁸ × 3.55 × 10⁻⁸

Kw = 1.3 × 10⁻¹⁵

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AveGali [126]

Answer:

18.075 mL of NaOH solution was added to achieve neutralization

Explanation:

First, let's formulate the chemical reaction between nitric acid and sodium hydroxide:

NaOH + HNO3 → NaNO3 + H2O

From this balanced equation we know that 1 mole of NaOH reacts with 1 mole of HNO3 to achieve neutralization. Let's calculate how many moles we have in the 25 mL aliquot to be titrated:

63.01 g of HNO3 ----- 1 mole

8.2 g of HNO3 ----- x = (8.2 g × 1 mole)/63.01 g = 0.13014 moles of HNO3

So far we added 8.2 grams of nitric acid (0.13014 moles) in 1 L of water.

1000 mL solution ---- 0.13014 moles of HNO3

25 mL (aliquot) ---- x = (25 mL× 0.13014 moles)/1000 mL = 0.0032535 moles

So, we now know that in the 25 mL aliquot to be titrated we have 0.0032535 moles of HNO3. As we stated before, 1 mole of NaOH will react with 1 mole of HNO3, hence 0.0032535 moles of HNO3 have to react with 0.0032535 moles of NaOH to achieve neutralization. Let's calculate then, in which volume of the given NaOH solution we have 0.0032535 moles:

0.18 moles of NaOH ----- 1000 mL Solution

0.0032535 moles---- x=(0.0032535moles×1000 mL)/0.18 moles = 18.075mL

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4 0
2 years ago
. Calculate the molality of each of the following solutions: (a) 0.710 kg of sodium carbonate (washing soda), Na2CO3, in 10.0 kg
Step2247 [10]

Answer:

a)0.67 molal

b) 5.67 molal

c) 2.82 molal

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Explanation:

Step 1: Data given

Molality = moles solute / mass solvent

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Calculate moles of Na2CO3 =  710 grams / 105.99 g/mol = 6.70 moles

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