4. Find the pH at each of the following points in the titration of 25 mL of 0.3 M HF with 0.3 M NaOH. The Ka value is 6.6x10-4 a. The initial pH b. After adding 10mL of 0.3 M NaOH c. After adding 12.5 mL of 0.3 M NaOH d. After adding 25 mL of 0.3 M NaOH e. After adding 26 mL of 0.3 M NaOH

Explanation:

Since HF is a weak acid, the use of an ICE table is required to find the pH. The question gives us the concentration of the HF.

HF+H2O⇌H3O++F−HF+H2O⇌H3O++F−

Initial0.3 M-0 M0 MChange- X-+ X+XEquilibrium0.3 - X-X MX M

Writing the information from the ICE Table in Equation form yields

6.6×10−4=x20.3−x6.6×10−4=x20.3−x

Manipulating the equation to get everything on one side yields

0=x2+6.6×10−4x−1.98×10−40=x2+6.6×10−4x−1.98×10−4

Now this information is plugged into the quadratic formula to give

x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√2x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)2

The quadratic formula yields that x=0.013745 and x=-0.014405

However we can rule out x=-0.014405 because there cannot be negative concentrations. Therefore to get the pH we plug the concentration of H3O+ into the equation pH=-log(0.013745) and get pH=1.86

n has four pairs of electrons in the valence shell, electrons repel one another ,n has 7 protons in its nucleus, and since n has 7 protons, it must fill the second shell, giving it 4 pairs of electrons. the electrons form 3 bonds and 1 lone pair of electrons. each pair of electrons repels the other pairs, so they are equally far apart.

so all of the above

where are the choice options?