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.
1
Expert's answer
2020-04-06T12:06:23-0400

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+]=(CKa)[H^+] = \sqrt (C * Ka)

[H+]=(0.2506.8104)=1.3102[H^+] = \sqrt (0.250 * 6.8*10^{-4}) = 1.3*10^{-2}

pH=log(H+)pH = -log(H^+)

pH=log(1.3102)=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=(1.3102/0.250)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+]=(0.1306.8104)=9.4103[H^+] = \sqrt (0.130 * 6.8*10^{-4}) = 9.4*10^{-3}

pH=log(9.4103)=2.03pH = -log(9.4*10^{-3}) = 2.03


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

A=(9.4103/0.130)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+]=(4.001026.8104)=5.2103[H^+] = \sqrt (4.00*10^{-2} * 6.8*10^{-4}) = 5.2*10^{-3}

pH=log(5.2103)=2.28pH = -log(5.2*10^{-3}) = 2.28


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

A=(5.2103/4.0102)100%=13%\Alpha = (5.2*10^{-3}/4.0*10^{-2})*100\% = 13\%


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