1. What is the pH of a 0.350 M solution of lithium citrate? (Ka Citric Acid = 7.4 x 10-4)

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1. What is the pH of a 0.350 M solution of lithium citrate? (Ka Citric Acid = 7.4 x 10-4)

 

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1. К, СH,О, — 7.4х 10 4
[Li,C,H¸0(ag] = 0.350M
pH =
Li,C„H,O(aq) + H,0÷ C,H¿0laq) +OH
(ag)
7(ag)
Li,C,H¿O(ag)
7(ag)
OH
(ag)
0.350M
+x
+x
E
0.350 -x
-14
Kw = 1.0 x 10
Ka = 7.4 x 10
K, = ?
K«K, = Kw
K, =
K
K.
= LOs10 -14
7.4x10
-11
= 1.4 x 10
[C,H,Oa[OH a
[Li,C,H5O)
K, =
-11
[C,H,0w[OH nel
1.4 x 10
1.4 x 10 -11
[x][x]
[0.350-x]
[Li,C,H;0rlap]
Rule of 100:
0.350
1.4x10
= 25900000000
25900000000 > 100
Therefore, 0.350 -x = 0.350
1.4 x 10 -11 .
x? = 4.7 x 10 -12
0.350
x= 2.175 x 10 6
Transcribed Image Text:1. К, СH,О, — 7.4х 10 4 [Li,C,H¸0(ag] = 0.350M pH = Li,C„H,O(aq) + H,0÷ C,H¿0laq) +OH (ag) 7(ag) Li,C,H¿O(ag) 7(ag) OH (ag) 0.350M +x +x E 0.350 -x -14 Kw = 1.0 x 10 Ka = 7.4 x 10 K, = ? K«K, = Kw K, = K K. = LOs10 -14 7.4x10 -11 = 1.4 x 10 [C,H,Oa[OH a [Li,C,H5O) K, = -11 [C,H,0w[OH nel 1.4 x 10 1.4 x 10 -11 [x][x] [0.350-x] [Li,C,H;0rlap] Rule of 100: 0.350 1.4x10 = 25900000000 25900000000 > 100 Therefore, 0.350 -x = 0.350 1.4 x 10 -11 . x? = 4.7 x 10 -12 0.350 x= 2.175 x 10 6
[OH (aq
] = x
= 2.175 x 10 -6
pH =- log[OH ag)]
=- log(2.175 x 10 )
= 5.66
pH + pOН — 14
pH - 14- рОН
= 14 – 5.66
= 8.34
Therefore, the pH of a 0.350M solution of lithium citrate is 8.34.
Transcribed Image Text:[OH (aq ] = x = 2.175 x 10 -6 pH =- log[OH ag)] =- log(2.175 x 10 ) = 5.66 pH + pOН — 14 pH - 14- рОН = 14 – 5.66 = 8.34 Therefore, the pH of a 0.350M solution of lithium citrate is 8.34.
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