A buffer is prepared by adding 24.6 mL of 0.65 M to 49.6 mL of 0.28 M. What is the pH of the final solution?

The given question is incomplete, the complete question is:

A buffer is prepared by adding 24.6 mL of 0.65 M NaF to 49.6 mL of 0.28 M HF. What is the pH of the final solution?

The correct answer is 3.23.

Explanation:

The calculations of pH in a buffer solution is determined with the help of Henderson-Hasselbalch equation, that is,

pH = pKa + log[conjugate base]/[acid]

= pKa + log[A⁻][HA]

Here Ka is the equilibrium constant.  

Based on the given information, the volume and concentration of NaF is 24.6 ml and 0.65 M. Now the millimoles of NaF can be determined by using the formula,  

millimoles = Molarity × Volume

= (0.65 M) (24.6 ml) (1 mmol/ml/1M)

= 16 mmol

Thus, the millimoles of NaF or F⁻ is 16 mmol.  

On the other hand, the volume and concentration given of HF is 49.6 ml and 0.28 M. Now the millimoles of HF will be,  

Millimoles = 0.28 M × 49.6 ml (1mmol/ml/1M)

Millimoles = 13.9 mmol

Thus, the millimoles of HF is 13.9 mmol.  

The value of Ka for strong acid is 6.8 × 10⁻⁴.  

Now the pH of the buffer solution can be determined by using the above mentioned Henderson-Hasselbalch equation,  

pH = pKa + log[F⁻]/[HF]

= -logKa + log[16 mmol]/[13.9 mmol]

= -log(6.8 × 10⁻⁴) + 0.06

= 3.17 + 0.06

= 3.23.  

Hence, the pH of the solution is 3.23.

the answer would be b i believe