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The pH of an acidic buffer mixture is:...

The `pH` of an acidic buffer mixture is:

A

`gt7`

B

`lt7`

C

`=7`

D

Depends upon `K_(a)` of acids

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To determine the pH of an acidic buffer mixture, we can use the Henderson-Hasselbalch equation, which is a fundamental equation in buffer chemistry. Here’s a step-by-step solution: ### Step 1: Understand the Henderson-Hasselbalch Equation The pH of an acidic buffer solution can be calculated using the following equation: \[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \] Where: - \(\text{pH}\) is the pH of the buffer solution. - \(\text{pK}_a\) is the negative logarithm of the acid dissociation constant (\(K_a\)) of the weak acid in the buffer. - \([\text{Salt}]\) is the concentration of the salt (the conjugate base). - \([\text{Acid}]\) is the concentration of the weak acid. ### Step 2: Identify the Components of the Buffer In an acidic buffer, you typically have a weak acid (HA) and its conjugate base (A⁻, usually in the form of a salt). For example, acetic acid (CH₃COOH) and sodium acetate (CH₃COONa) can form an acidic buffer. ### Step 3: Determine the pK_a Value The pK_a value is specific to the weak acid used in the buffer. You can find this value in tables of acid dissociation constants. For acetic acid, the pK_a is approximately 4.76. ### Step 4: Calculate the Ratio of Salt to Acid Concentration To use the Henderson-Hasselbalch equation, you need the concentrations of the salt and the acid. If you have these values, you can calculate the ratio \(\frac{[\text{Salt}]}{[\text{Acid}]}\). ### Step 5: Substitute Values into the Equation Once you have the pK_a and the ratio of salt to acid, substitute these values into the Henderson-Hasselbalch equation to find the pH. ### Example Calculation If you have a buffer solution with: - \([\text{Salt}] = 0.1 \, M\) - \([\text{Acid}] = 0.05 \, M\) - \(\text{pK}_a = 4.76\) You would calculate: \[ \text{pH} = 4.76 + \log\left(\frac{0.1}{0.05}\right) \] \[ \text{pH} = 4.76 + \log(2) \] \[ \text{pH} = 4.76 + 0.301 \] \[ \text{pH} \approx 5.06 \] ### Conclusion The pH of the acidic buffer mixture depends on the pK_a of the weak acid and the ratio of the concentrations of the salt and the acid. ---

To determine the pH of an acidic buffer mixture, we can use the Henderson-Hasselbalch equation, which is a fundamental equation in buffer chemistry. Here’s a step-by-step solution: ### Step 1: Understand the Henderson-Hasselbalch Equation The pH of an acidic buffer solution can be calculated using the following equation: \[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \] Where: - \(\text{pH}\) is the pH of the buffer solution. - \(\text{pK}_a\) is the negative logarithm of the acid dissociation constant (\(K_a\)) of the weak acid in the buffer. ...
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The pH of basic buffer mixtures is given by : pH=pK_(a)+log((["Base"])/(["Salt"])) , whereas pH of acidic buffer mixtures is given by: pH= pK_(a)+log((["Salt"])/(["Acid"])) . Addition of little acid or base although shows no appreciable change for all practical purpose, but since the ratio (["Base"])/(["Salt"]) or (["Salt"])/(["Acid"]) change, a slight decrease or increase in pH results in. Mole of HCI required to prepare a buffer solution of pH=8.5 with 0.1 mole of NaCN in one litre solution is: (pK_(a) for CN^(-)=4.61)

The pH of basic buffer mixtures is given by : pH=pK_(a)+log((["Base"])/(["Salt"])) , whereas pH of acidic buffer mixtures is given by: pH= pK_(a)+log((["Salt"])/(["Acid"])) . Addition of little acid or base although shows no appreciable change for all practical purpose, but since the ratio (["Base"])/(["Salt"]) or (["Salt"])/(["Acid"]) change, a slight decrease or increase in pH results in. The volume of 0.2M NaOH needed to prepare a buffer of pH 4.74 with 50 mL of 0.2M acetic acid is: (pK_(a) of CH_(3)COO^(-)= 9.26)

The pH of basic buffer mixtures is given by : pH=pK_(a)+log((["Base"])/(["Salt"])) , whereas pH of acidic buffer mixtures is given by: pH= pK_(a)+log((["Salt"])/(["Acid"])) . Addition of little acid or base although shows no appreciable change for all practical purpose, but since the ratio (["Base"])/(["Salt"]) or (["Salt"])/(["Acid"]) change, a slight decrease or increase in pH results in. The amount of (NH_(4))_(2)SO_(4) to be added to 500 mL of 0.01M NH_(4)OH solution (pK_(a) for NH_(4)^(+) is 9.26) prepare a buffer of pH 8.26 is:

The pH of basic buffer mixtures is given by : pH=pK_(a)+log((["Base"])/(["Salt"])) , whereas pH of acidic buffer mixtures is given by: pH= pK_(a)+log((["Salt"])/(["Acid"])) . Addition of little acid or base although shows no appreciable change for all practical purpose, but since the ratio (["Base"])/(["Salt"]) or (["Salt"])/(["Acid"]) change, a slight decrease or increase in pH results in. A solution containing 0.2 mole of dichloroacetic acid (K_(a)=5xx10^(-2)) and 0.1 mole sodium dichloroacetate in one litre solution has [H^(+)] :

The pH of basic buffer mixtures is given by : pH=pK_(a)+log((["Base"])/(["Salt"])) , whereas pH of acidic buffer mixtures is given by: pH= pK_(a)+log((["Salt"])/(["Acid"])) . Addition of little acid or base although shows no appreciable change for all practical purpose, but since the ratio (["Base"])/(["Salt"]) or (["Salt"])/(["Acid"]) change, a slight decrease or increase in pH results in. A weak acid HA after treatment with 12 mL of 0.1M strong base BOH has a pH=5 . At end point, the volume of same base required is 26.6mL K_(a) of acid is:

The Ph of basic buffer mixtures is given by : Ph=Pk_(a)+ log (["Base"])/(["Salt"]) whereas Ph of acidic buffer mixtures is given by : Ph = pK_(a)+"log"(["Salt"])/(["Acid"]) . Addition of little acid or base although shows no appreciable change in Ph for all practical purposes, but sicne the ratio (["Base"])/(["Salt"]) or (["Salt"])/(["Acid"]) changes, a slight decrease or increase in pH results. The volume of 0.2 m NaOH needed to prepare a buffer of pH 4.74 with 50 mL of 0.2 m acetic acid pH_(b) of CH_(3)COO^(-)=9.26 is :

A2Z-IONIC EQUILIBIUM-Buffer Solutions
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  2. Which may be added to one litre of water to act a buffer?

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  3. pH of a solution of 10 ml. 1 N sodium acetate and 50 ml 2N acetic acid...

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  4. The pH of an acidic buffer mixture is:

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  5. pH of a mixture containing 0.10 M X^(-) and 0.20 M HX is: [pK(b)(X^(-)...

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  6. What % of the carbon in the H(2)CO(3)-HCO(3)^(-) buffer should be in t...

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  7. An acidic buffer solution can be prepared by mixing solution of

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  8. Buffer capacity of a buffer solution is x, the volume of 1 M NaOH adde...

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  9. 100 mL of a buffer solution contains 0.1 M each of weak acid HA and sa...

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  10. Calculate the amount of (NH(4))(2)SO(4) in grams which must be added t...

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  11. The buffer solution of 100 ml having a pH value 4 when added to 1 ml d...

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  12. What amount of solution propanoate should be added to one litre of an ...

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  13. 20ml of 0.2 M NaOH is added to 50 ml, of 0.2 M CH(3)COOH to give 70 ml...

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  14. The concentration of H^(+) ion in a 0.2 M solution of HCOOH is 6.4xx10...

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  15. What amount of HCl will be required to prepare one litre of a buffer s...

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  16. Calculate pH of a solution of given mixture (0.1 "mol " CH(3)COOH+0.2 ...

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  17. Calculate the pH of a buffer solution prepared by dissolving 10.6 g of...

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  20. The pK(a) of a weak acid (HA) is 4.5. The pOH of an aqueous buffered s...

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