The answer is going to be A.
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
Answer:
D
Explanation:
To answer this question, we will need to write the dissociation equation of aluminum trichloride.
AlCl3 ——-> Al3+ + 3Cl-
It can be seen that when aluminum chloride dissociates, it gives one mole of aluminum ion and three moles of the chloride ion.
From here we can see that the concentration of the aluminum chloride equals that of the aluminum ion while that of the chloride ion is thrice that of the aluminum chloride. This means we simply multiply 0.12M by 3 to get the molarity of the chloride ion while that of the aluminum ion remains the same
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