A 35.00 mL solution of 0.2500 M #HF# is titrated with a standardized 0.1532 M solution of #NaOH# at 25 degrees C. What is the pH of the #HF# solution before titrant is added?
For #HF#, the
Ka = 6.8 x 10–4 . W

Question

A 35.00 mL solution of 0.2500 M #HF# is titrated with a standardized 0.1532 M solution of #NaOH# at 25 degrees C.  What is the pH of the #HF# solution before titrant is added?  
For #HF#, the  
Ka = 6.8 x 10–4 .  W

Naomi Aronson · Accepted Answer

#1.88#

Using the following phrase is a short cut to accomplish this:

#pH=1/2(pK_a-loga)#

Where #a# is the concentration of the weak acid.

#K_a=6.8xx10^-4" " "mol/l"#

#:.pK_a=-log[6.8xx10^-4]=3.167#

#:.pH =1/2[3.167-log(0.25)]#

#pH=1/2[3.167-(-0.6)]=1.88#

Kai Bowman · Answer

undefined To find the pH of the HF solution before titrant is added, you need to calculate the concentration of HF after mixing with water. Then, use the given Ka value to find the concentration of H+ ions. Finally, use the concentration of H+ ions to calculate the pH using the formula pH = -log[H+].

Wyatt Adkins · Answer

undefined To find the pH of the HF solution before titrant is added, you can use the equilibrium expression for the dissociation of HF:

$$ K_a = \frac{{[\text{H}^+][\text{F}^-]}}{{[\text{HF}]}} $$

Given that the initial concentration of HF is 0.2500 M, and assuming that all of the HF dissociates, the concentration of $[\text{HF}]$ after dissociation would be $(0.2500 - x)$ M, where $x$ is the concentration of $[\text{H}^+]$ and $[\text{F}^-]$ ions.

Since $x$ is small compared to 0.2500 M, we can approximate $(0.2500 - x)$ as 0.2500 M.

Plugging the values into the equilibrium expression and solving for $x$, we get:

$$ 6.8 \times 10^{-4} = \frac{{x^2}}{{0.2500}} $$

$$ x = \sqrt{{(6.8 \times 10^{-4}) \times 0.2500}} $$

$$ x = 0.0052 $$

Thus, the concentration of $[\text{H}^+]$ ions in the HF solution before titrant is added is approximately $0.0052 \, \text{M}$.

To find the pH, use the formula:

$$ \text{pH} = -\log{[\text{H}^+]} $$

$$ \text{pH} = -\log{(0.0052)} $$

$$ \text{pH} \approx 2.28 $$

A 35.00 mL solution of 0.2500 M #HF# is titrated with a standardized 0.1532 M solution of #NaOH# at 25 degrees C. What is the pH of the #HF# solution before titrant is added? For #HF#, the Ka = 6.8 x 10–4 . W