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Natasha_Volkova [10]
3 years ago
12

The pOH of an aqueous solution at 25°C was found to be 1.20. The pH of this solution is . The hydronium ion concentration is M.

The hydroxide ion concentration is M.
Chemistry
1 answer:
frosja888 [35]3 years ago
5 0

Answer:

pH = 12.80

[H3O+] = 1.58 * 10^-13 M

[OH-] = 0.063 M

Explanation:

Step 1: Data given

pOH = 1.20

Temperature = 25.0 °C

Step 2: Calulate pH

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 1.20 = 12.80

Step 3: Calculate hydronium ion concentration

pH = -log[H+] = -log[H3O+]

12.80 = -log[H3O+]

10^-12.80 = [H3O+] = 1.58 * 10^-13 M

Step 4: Calculate the hydroxide ion concentration

pOH = 1.20 = -log [OH-]

10^-1.20 = [OH-] = 0.063M

Step 5: Control [H3O+] and [OH-]

[H3O+]*[OH-] = 1* 10^-14

1.58 *10^-13 * 0.063 = 1* 10^-14

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Which of the following pairs of elements could possibly be in the same group? X has a 1+ ion; Y has a 1- ion. X tends to form a
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How many milliliters of 0.150 M NaOH are required to neutralize 85.0 mL of 0.300 M H2SO4 ? The balanced neutralization reaction
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Answer : The volume of NaOH required to neutralize is, 340 mL

Explanation :

To calculate the volume of base (NaOH), we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=0.300M\\V_1=85.0mL\\n_2=1\\M_2=0.150M\\V_2=?

Putting values in above equation, we get:

2\times 0.300M\times 85.0mL=1\times 0.150M\times V_2\\\\V_2=340mL

Hence, the volume of NaOH required to neutralize is, 340 mL

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3 years ago
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