Answer :
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
When the pH we plug the concentration of H3O+ into the equation pH is =-log(0.013745) and also get pH=1.86
What is the Weak Acid?
When the HF is a weak acid, Then the use of an ICE table is required to find the pH. The question that gives us the concentration of the HF is:
HF+H2O⇌H3O++F−HF+H2O⇌H3O++F−
Now, The Initial0.3 M-0 M0 MChange- X-+ X+XEquilibrium0.3 - X-X MX M
After that, Writing the information from the ICE Table in Equation form yields
Then, 6.6×10−4=x20.3−x6.6×10−4=x20.3−x
Now, Manipulating the equation to get everything on one side yields
Also, 0 is =x2+6.6×10−4x−1.98×10−40=x2+6.6×10−4x−1.98×10−4
Now this the information is plugged into the quadratic formula to give
Then x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√2x is =−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)2
Then The quadratic formula yields that x=0.013745 and also x is =-0.014405
However, when we can rule out x is =-0.014405 because there can't be negative concentrations.
Thus to get the pH we plug the concentration of H3O+ into the equation pH is =-log(0.013745) and also get pH=1.86
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