Home
Class 12
CHEMISTRY
The pH of the solution produced by mixin...

The pH of the solution produced by mixing equal volume of `2.0 xx 10^(-3) M HClO_(4)` and `1.0 xx 10^(-2) M KClO_(4)` is

A

2.7

B

2.3

C

3

D

1

Text Solution

AI Generated Solution

The correct Answer is:
To find the pH of the solution produced by mixing equal volumes of `2.0 x 10^(-3) M HClO4` and `1.0 x 10^(-2) M KClO4`, we can follow these steps: ### Step 1: Understand the Components - **HClO4** is a strong acid and will completely dissociate in solution to give H⁺ ions. - **KClO4** is a salt that will dissociate into K⁺ and ClO4⁻ ions, but it does not affect the pH since ClO4⁻ is the conjugate base of a strong acid and does not hydrolyze. ### Step 2: Calculate the Initial Moles Assuming we mix equal volumes of both solutions, let's say we mix 1 liter of each solution for simplicity: - Moles of HClO4 = Volume × Concentration = 1 L × `2.0 x 10^(-3) M` = `2.0 x 10^(-3) moles` - Moles of KClO4 = Volume × Concentration = 1 L × `1.0 x 10^(-2) M` = `1.0 x 10^(-2) moles` ### Step 3: Calculate the Total Volume The total volume after mixing will be: - Total Volume = Volume of HClO4 + Volume of KClO4 = 1 L + 1 L = 2 L ### Step 4: Calculate the New Concentration of HClO4 Since HClO4 completely dissociates, the concentration of H⁺ ions will be equal to the concentration of HClO4 after mixing: - New concentration of HClO4 = Moles of HClO4 / Total Volume = `(2.0 x 10^(-3) moles) / (2 L)` = `1.0 x 10^(-3) M` ### Step 5: Calculate the pH The concentration of H⁺ ions is `1.0 x 10^(-3) M`. The pH can be calculated using the formula: - pH = -log[H⁺] - pH = -log(1.0 x 10^(-3)) = 3.0 ### Final Answer The pH of the solution produced by mixing equal volumes of `2.0 x 10^(-3) M HClO4` and `1.0 x 10^(-2) M KClO4` is **3.0**. ---
To find the pH of the solution produced by mixing equal volumes of `2.0 x 10^(-3) M HClO4` and `1.0 x 10^(-2) M KClO4`, we can follow these steps: ### Step 1: Understand the Components - **HClO4** is a strong acid and will completely dissociate in solution to give H⁺ ions. - **KClO4** is a salt that will dissociate into K⁺ and ClO4⁻ ions, but it does not affect the pH since ClO4⁻ is the conjugate base of a strong acid and does not hydrolyze. ### Step 2: Calculate the Initial Moles Assuming we mix equal volumes of both solutions, let's say we mix 1 liter of each solution for simplicity: ...
Promotional Banner

Similar Questions

Explore conceptually related problems

The pH of the solutions produced by mixing equal volumes of 2.0 xx 10^(-3) M HClO_(4) and 1.0 xx 10^(-2) MKClO_(4) is

Given that solubility product of BaSO_(4) is 1 xx 10^(-10) will be precipiate from when a. Equal volumes of 2 xx 10^(-3)M BaC1_(2) solution and 2 xx 10^(-4)M Na_(2)SO_(4) solution, are mixed? b. Equal volumes of 2 xx 10^(-8) M BaC1_(2) solution and 2 xx 10^(-3)M Na_(2)SO_(4) solution, are mixed? c. 100mL of 10^(-3)M BaC1_(2) and 400mL of 10^(-6)M Na_(2)SO_(4) are mixed.

pH of solution obtained by mixing equal volumes of 0.1M Triethyl amine (K_(b)=6.4xx10^(-5)) & (4)/(45)M NH_(4)OH (K_(b)=1.8xx10^(-5)) will be :

Give reason why BaSO_(4) will precipitate out when equal volumes of 2xx10^(-3) M BaCl_(2) solution and 2xx10^(-4)M Na_(2)SO_(4) solution are mixed. Given that the solubility product of BaSO_(4) is 1xx10^(-10).

Calculate the pH of the following solutions obtained by mixing : (a)equal volumes of two solutions with pH=4 and pH=10 (b)400 mL of solution with pH=3 and 100 mL of solution with pH=4 (c )200 mL of 0.1 M NaOH and 300 mL of 0.2 KOH

The pH of a solution obtained by mixing 50 mL of 1 M HCl and 30 mL of 1M NaOH is x xx 10^(-4) . The value of x is ________. (Nearest inetger) [log 2.5 = 0.3979]