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 ∗ K a ) [H^+] = \sqrt (C * Ka) [ H + ] = ( C ∗ K a )
[ H + ] = ( 0.250 ∗ 6.8 ∗ 1 0 − 4 ) = 1.3 ∗ 1 0 − 2 [H^+] = \sqrt (0.250 * 6.8*10^{-4}) = 1.3*10^{-2} [ H + ] = ( 0.250 ∗ 6.8 ∗ 1 0 − 4 ) = 1.3 ∗ 1 0 − 2
p H = − l o g ( H + ) pH = -log(H^+) p H = − l o g ( H + )
p H = − l o g ( 1.3 ∗ 1 0 − 2 ) = 1.9 pH = -log(1.3*10^{-2}) = 1.9 p H = − l o g ( 1.3 ∗ 1 0 − 2 ) = 1.9
2) Find the percent dissociation of a 0.250 M HF solution.
A = ( [ H + ] / [ H F ] ) ∗ 100 % \Alpha = ([H^+]/[HF])*100\% A = ([ H + ] / [ H F ]) ∗ 100%
A = ( 1.3 ∗ 1 0 − 2 / 0.250 ) ∗ 100 % = 5.2 % \Alpha = (1.3*10^{-2}/0.250)*100\% = 5.2\% A = ( 1.3 ∗ 1 0 − 2 /0.250 ) ∗ 100% = 5.2%
3) Find the pH of a 0.130 M HF solution.
[ H + ] = ( 0.130 ∗ 6.8 ∗ 1 0 − 4 ) = 9.4 ∗ 1 0 − 3 [H^+] = \sqrt (0.130 * 6.8*10^{-4}) = 9.4*10^{-3} [ H + ] = ( 0.130 ∗ 6.8 ∗ 1 0 − 4 ) = 9.4 ∗ 1 0 − 3
p H = − l o g ( 9.4 ∗ 1 0 − 3 ) = 2.03 pH = -log(9.4*10^{-3}) = 2.03 p H = − l o g ( 9.4 ∗ 1 0 − 3 ) = 2.03
4) Find the percent dissociation of a 0.130 M HF solution.
A = ( 9.4 ∗ 1 0 − 3 / 0.130 ) ∗ 100 % = 7.2 % \Alpha = (9.4*10^{-3}/0.130)*100\% = 7.2\% A = ( 9.4 ∗ 1 0 − 3 /0.130 ) ∗ 100% = 7.2%
5) Find the pH of a 4.00×10−2 M HF solution.
[ H + ] = ( 4.00 ∗ 1 0 − 2 ∗ 6.8 ∗ 1 0 − 4 ) = 5.2 ∗ 1 0 − 3 [H^+] = \sqrt (4.00*10^{-2} * 6.8*10^{-4}) = 5.2*10^{-3} [ H + ] = ( 4.00 ∗ 1 0 − 2 ∗ 6.8 ∗ 1 0 − 4 ) = 5.2 ∗ 1 0 − 3
p H = − l o g ( 5.2 ∗ 1 0 − 3 ) = 2.28 pH = -log(5.2*10^{-3}) = 2.28 p H = − l o g ( 5.2 ∗ 1 0 − 3 ) = 2.28
6) Find the percent dissociation of a 4.00×10−2 M HF solution.
A = ( 5.2 ∗ 1 0 − 3 / 4.0 ∗ 1 0 − 2 ) ∗ 100 % = 13 % \Alpha = (5.2*10^{-3}/4.0*10^{-2})*100\% = 13\% A = ( 5.2 ∗ 1 0 − 3 /4.0 ∗ 1 0 − 2 ) ∗ 100% = 13%