Question #107419

Find the pH and percent ionization of each HF solution. (Ka for HF is 6.8×10^−4.)


1) Find the pH of a 0.250 M HF solution.

2) Find the percent dissociation of a 0.250 M HF solution.

3) Find the pH of a 0.130 M HF solution.

4) Find the percent dissociation of a 0.130 M HF solution.

5) Find the pH of a 4.00×10−2 M HF solution.

6) Find the percent dissociation of a 4.00×10−2 M HF solution.

Expert's answer

Find the pH and percent ionization of each HF solution. (Ka for HF is 6.8×10^−4.)

1) Find the pH of a 0.250 M HF solution

[H+]=(C∗Ka)[H^+] = \sqrt (C * Ka)

[H+]=(0.250∗6.8∗10−4)=1.3∗10−2[H^+] = \sqrt (0.250 * 6.8*10^{-4}) = 1.3*10^{-2}

pH=−log(H+)pH = -log(H^+)

pH=−log(1.3∗10−2)=1.9pH = -log(1.3*10^{-2}) = 1.9


2) Find the percent dissociation of a 0.250 M HF solution.

A=([H+]/[HF])∗100%\Alpha = ([H^+]/[HF])*100\%

A=(1.3∗10−2/0.250)∗100%=5.2%\Alpha = (1.3*10^{-2}/0.250)*100\% = 5.2\%


3) Find the pH of a 0.130 M HF solution.

[H+]=(0.130∗6.8∗10−4)=9.4∗10−3[H^+] = \sqrt (0.130 * 6.8*10^{-4}) = 9.4*10^{-3}

pH=−log(9.4∗10−3)=2.03pH = -log(9.4*10^{-3}) = 2.03


4) Find the percent dissociation of a 0.130 M HF solution.

A=(9.4∗10−3/0.130)∗100%=7.2%\Alpha = (9.4*10^{-3}/0.130)*100\% = 7.2\%


5) Find the pH of a 4.00×10−2 M HF solution.

[H+]=(4.00∗10−2∗6.8∗10−4)=5.2∗10−3[H^+] = \sqrt (4.00*10^{-2} * 6.8*10^{-4}) = 5.2*10^{-3}

pH=−log(5.2∗10−3)=2.28pH = -log(5.2*10^{-3}) = 2.28


6) Find the percent dissociation of a 4.00×10−2 M HF solution.

A=(5.2∗10−3/4.0∗10−2)∗100%=13%\Alpha = (5.2*10^{-3}/4.0*10^{-2})*100\% = 13\%


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