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When a salt reacts with water to form a...

When a salt reacts with water to form acidic or basic solution , the process is called hydrolysis . The pH of salt solution can be calculated using the following relations :
`pH = 1/2 [pK_(w) +pK_(a) + logc] ` (for salt of weak acid and strong base .)
`pH = 1/2 [pK_(w) - pK_(b) - logc] ` (for salt of weak base and strong acid ) .
`pH = 1/2 [ pK_(w)+pK_(a)-pK_(b)] ` (for weak acid and weak base ).
where 'c' represents the concentration of salt .
When a weak acid or a weak base not completely neutralized by strong base or strong acid
respectively , then formation of buffer takes place . The pH of buffer solution can be calculated using the following relation :
`pH = pK_(a) + log . (["Salt"])/(["Acid"]) , pOH = pK_(b) + log . (["Salt"])/([ "Base"])`
Answer the following questions using the following data :
`pK_(a) = 4.7447 , pK_(b) = 4.7447 ,pK_(w) = 14`
`0.001` M `NH_(4)Cl` aqueous solution has pH :

A

`6.127`

B

`7.126`

C

`2.167`

D

`1.267`

Text Solution

AI Generated Solution

The correct Answer is:
To find the pH of a 0.001 M NH4Cl aqueous solution, we will follow these steps: ### Step 1: Identify the type of salt NH4Cl is a salt formed from the strong acid HCl and the weak base NH4OH. Therefore, we can use the formula for calculating the pH of a salt of a strong acid and a weak base. ### Step 2: Write the formula for pH For a salt of a strong acid and a weak base, the pH can be calculated using the formula: \[ \text{pH} = \frac{1}{2} \left( pK_w - pK_b - \log c \right) \] ### Step 3: Substitute the known values We are given: - \( pK_w = 14 \) - \( pK_b = 4.7447 \) - \( c = 0.001 \, M \) Now substituting these values into the formula: \[ \text{pH} = \frac{1}{2} \left( 14 - 4.7447 - \log(0.001) \right) \] ### Step 4: Calculate \(\log(0.001)\) \[ \log(0.001) = \log(10^{-3}) = -3 \] ### Step 5: Substitute \(\log(0.001)\) into the pH formula Now substituting \(-3\) into the equation: \[ \text{pH} = \frac{1}{2} \left( 14 - 4.7447 + 3 \right) \] ### Step 6: Simplify the equation Calculating inside the parentheses: \[ 14 - 4.7447 + 3 = 12.2553 \] ### Step 7: Final calculation of pH Now divide by 2: \[ \text{pH} = \frac{12.2553}{2} = 6.12765 \] ### Step 8: Round to appropriate significant figures Thus, the pH of the solution is approximately: \[ \text{pH} \approx 6.127 \] ### Final Answer The pH of the 0.001 M NH4Cl aqueous solution is **6.127**. ---
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When a salt reacts with water to form acidic or basic solution , the process is called hydrolysis . The pH of salt solution can be calculated using the following relations : pH = 1/2 [pK_(w) +pK_(a) + logc] (for salt of weak acid and strong base .) pH = 1/2 [pK_(w) - pK_(b) - logc] (for salt of weak base and strong acid ) . pH = 1/2 [ pK_(w)+pK_(a)-pK_(b)] (for weak acid and weak base ). where 'c' represents the concentration of salt . When a weak acid or a weak base not completely neutralized by strong base or strong acid respectively , then formation of buffer takes place . The pH of buffer solution can be calculated using the following relation : pH = pK_(a) + log . (["Salt"])/(["Acid"]) , pOH = pK_(b) + log . (["Salt"])/([ "Base"]) Answer the following questions using the following data : pK_(a) = 4.7447 , pK_(b) = 4.7447 ,pK_(w) = 14 When 100 mL of 0.1 M NH_(4)OH is added to 50 mL of 0.1M HCl solution , the pH is

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