QuestionGiven $\mathrm{pH}=9.3$ for a $\mathrm{KOH}$ solution, find the true statement among the options provided. Also, which is neutralized by $4 \mathrm{~mol}$ of $\mathrm{NaOH}$ ?

Studdy Solution

STEP 1

Assumptions1. The pH of the KOH solution is9.3. The relationship between pH, pOH, and the concentrations of $\mathrm{OH}^-$ and $\mathrm{H}_3\mathrm{}^+$ are governed by the following equations a. $\mathrm{pH} + \mathrm{pOH} =14$ b. $\mathrm{pH} = -\log[\mathrm{H}_3\mathrm{}^+]$ c. $\mathrm{pOH} = -\log[\mathrm{OH}^-]$
3. The neutralization reaction between NaOH and an acid is a11 stoichiometric reaction.

STEP 2

First, we need to find the pOH of the solution using the relationship between pH and pOH.
$\mathrm{pOH} =14 - \mathrm{pH}$

STEP 3

Now, plug in the given value for the pH to calculate the pOH.
$\mathrm{pOH} =14 -9.3$

STEP 4

Calculate the pOH.
$\mathrm{pOH} =14 -9.3 =4.7$

STEP 5

Now, we can calculate the concentration of $\mathrm{OH}^-$ using the relationship between pOH and $\mathrm{OH}^-$ .
$[\mathrm{OH}^-] =10^{-\mathrm{pOH}}$

STEP 6

Plug in the calculated value for the pOH to find the concentration of $\mathrm{OH}^-$ .
$[\mathrm{OH}^-] =10^{-4.}$

STEP 7

Calculate the concentration of $\mathrm{OH}^-$ .
$[\mathrm{OH}^-] \approx2 \times10^{-5}$

STEP 8

Now, we can calculate the concentration of $\mathrm{H}_3\mathrm{}^+$ using the relationship between pH and $\mathrm{H}_3\mathrm{}^+$ .
$[\mathrm{H}_3\mathrm{}^+] =10^{-\mathrm{pH}}$

STEP 9

Plug in the given value for the pH to find the concentration of $\mathrm{H}_3\mathrm{}^+$ .
$[\mathrm{H}_3\mathrm{}^+] =^{-9.3}$

STEP 10

Calculate the concentration of $\mathrm{H}_3\mathrm{}^+$ .
$[\mathrm{H}_3\mathrm{}^+] \approx5 \times10^{-10}$ From the calculations, we can see thata. $\mathrm{pOH}=4.7$ (not5.7)
b. $[\mathrm{OH}^-]=2 \times10^{-5}$ (not $5 \times10^{-10}$ )
c. $\left[\mathrm{H}_{3} \mathrm{}^{+}\right]=5 \times10^{-10}$ (not $2 \times10^{-5}$ )
d. $\left[\mathrm{H}_{3} \mathrm{}^{+}\right]=5 \times10^{-10}$ (correct)
So, the correct answer is d.

STEP 11

Now, we need to find which acid is neutralized by4 moles of NaOH.Since the neutralization reaction between NaOH and an acid is a stoichiometric reaction,4 moles of NaOH will neutralize4 moles of any monoprotic acid.

Was this helpful?

QuestionGiven $\mathrm{pH}=9.3$ for a $\mathrm{KOH}$ solution, find the true statement among the options provided. Also, which is neutralized by $4 \mathrm{~mol}$ of $\mathrm{NaOH}$ ?

Studdy Solution

STEP 1

STEP 2

First, we need to find the pOH of the solution using the relationship between pH and pOH.
$\mathrm{pOH} =14 - \mathrm{pH}$

STEP 3

Now, plug in the given value for the pH to calculate the pOH.
$\mathrm{pOH} =14 -9.3$

STEP 4

Calculate the pOH.
$\mathrm{pOH} =14 -9.3 =4.7$

STEP 5

Now, we can calculate the concentration of $\mathrm{OH}^-$ using the relationship between pOH and $\mathrm{OH}^-$ .
$[\mathrm{OH}^-] =10^{-\mathrm{pOH}}$

STEP 6

Plug in the calculated value for the pOH to find the concentration of $\mathrm{OH}^-$ .
$[\mathrm{OH}^-] =10^{-4.}$

STEP 7

Calculate the concentration of $\mathrm{OH}^-$ .
$[\mathrm{OH}^-] \approx2 \times10^{-5}$

STEP 8

Now, we can calculate the concentration of $\mathrm{H}_3\mathrm{}^+$ using the relationship between pH and $\mathrm{H}_3\mathrm{}^+$ .
$[\mathrm{H}_3\mathrm{}^+] =10^{-\mathrm{pH}}$

STEP 9

Plug in the given value for the pH to find the concentration of $\mathrm{H}_3\mathrm{}^+$ .
$[\mathrm{H}_3\mathrm{}^+] =^{-9.3}$

STEP 10

STEP 11

Now, we need to find which acid is neutralized by4 moles of NaOH.Since the neutralization reaction between NaOH and an acid is a stoichiometric reaction,4 moles of NaOH will neutralize4 moles of any monoprotic acid.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionGiven pH=9.3\mathrm{pH}=9.3pH=9.3 for a KOH\mathrm{KOH}KOH solution, find the true statement among the options provided. Also, which is neutralized by 4 mol4 \mathrm{~mol}4 mol of NaOH\mathrm{NaOH}NaOH?

Studdy Solution

STEP 1

STEP 2

First, we need to find the pOH of the solution using the relationship between pH and pOH.pOH=14−pH\mathrm{pOH} =14 - \mathrm{pH}pOH=14−pH

STEP 3

Now, plug in the given value for the pH to calculate the pOH.pOH=14−9.3\mathrm{pOH} =14 -9.3pOH=14−9.3

STEP 4

Calculate the pOH.pOH=14−9.3=4.7\mathrm{pOH} =14 -9.3 =4.7pOH=14−9.3=4.7

STEP 5

Now, we can calculate the concentration of OH−\mathrm{OH}^-OH− using the relationship between pOH and OH−\mathrm{OH}^-OH−.[OH−]=10−pOH[\mathrm{OH}^-] =10^{-\mathrm{pOH}}[OH−]=10−pOH

STEP 6

Plug in the calculated value for the pOH to find the concentration of OH−\mathrm{OH}^-OH−.[OH−]=10−4.[\mathrm{OH}^-] =10^{-4.}[OH−]=10−4.

STEP 7

Calculate the concentration of OH−\mathrm{OH}^-OH−.[OH−]≈2×10−5[\mathrm{OH}^-] \approx2 \times10^{-5}[OH−]≈2×10−5

STEP 8

Now, we can calculate the concentration of H3+\mathrm{H}_3\mathrm{}^+H3​+ using the relationship between pH and H3+\mathrm{H}_3\mathrm{}^+H3​+.[H3+]=10−pH[\mathrm{H}_3\mathrm{}^+] =10^{-\mathrm{pH}}[H3​+]=10−pH

STEP 9

Plug in the given value for the pH to find the concentration of H3+\mathrm{H}_3\mathrm{}^+H3​+.[H3+]=−9.3[\mathrm{H}_3\mathrm{}^+] =^{-9.3}[H3​+]=−9.3

STEP 10

STEP 11

Now, we need to find which acid is neutralized by4 moles of NaOH.Since the neutralization reaction between NaOH and an acid is a stoichiometric reaction,4 moles of NaOH will neutralize4 moles of any monoprotic acid.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionGiven $\mathrm{pH}=9.3$ for a $\mathrm{KOH}$ solution, find the true statement among the options provided. Also, which is neutralized by $4 \mathrm{~mol}$ of $\mathrm{NaOH}$ ?

First, we need to find the pOH of the solution using the relationship between pH and pOH.
$\mathrm{pOH} =14 - \mathrm{pH}$

Now, plug in the given value for the pH to calculate the pOH.
$\mathrm{pOH} =14 -9.3$

Calculate the pOH.
$\mathrm{pOH} =14 -9.3 =4.7$

Now, we can calculate the concentration of $\mathrm{OH}^-$ using the relationship between pOH and $\mathrm{OH}^-$ .
$[\mathrm{OH}^-] =10^{-\mathrm{pOH}}$

Plug in the calculated value for the pOH to find the concentration of $\mathrm{OH}^-$ .
$[\mathrm{OH}^-] =10^{-4.}$

Calculate the concentration of $\mathrm{OH}^-$ .
$[\mathrm{OH}^-] \approx2 \times10^{-5}$

Now, we can calculate the concentration of $\mathrm{H}_3\mathrm{}^+$ using the relationship between pH and $\mathrm{H}_3\mathrm{}^+$ .
$[\mathrm{H}_3\mathrm{}^+] =10^{-\mathrm{pH}}$

Plug in the given value for the pH to find the concentration of $\mathrm{H}_3\mathrm{}^+$ .
$[\mathrm{H}_3\mathrm{}^+] =^{-9.3}$