Calculate the pH obtained by mixing 400mL of .2 M NaOH with 150mL of .1 M H3PO4 (pka's 2.12, 7.21, 12.32)
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There are 400 x 0.2 = 80 mmole of NaOH and 150 x0.1 = 15 mmole of H3PO4
comlete reaction between these will consume all the acid but only 45 mmole of NaOH
NaOH left = 35 mmole Since it is a strong base [OH-] from it will decide the pH with no contribution of H3PO4.
New [NaOH] = [OH-] = 35 /(400+ 150 = 550) = 6.36 x 10^-2
and pOH = - log 6.36x10^-2 = 1.1965
pH = 14 -pOH = 14-1.1965 = 12.8035
Pka are not required
comlete reaction between these will consume all the acid but only 45 mmole of NaOH
NaOH left = 35 mmole Since it is a strong base [OH-] from it will decide the pH with no contribution of H3PO4.
New [NaOH] = [OH-] = 35 /(400+ 150 = 550) = 6.36 x 10^-2
and pOH = - log 6.36x10^-2 = 1.1965
pH = 14 -pOH = 14-1.1965 = 12.8035
Pka are not required
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the majority of the H+ from H3PO4 will come in the first dissociation. H3PO4 --> H+ + H2PO4 -
the other 2 dissociations will produce H+ but it will be negligible to the pH
we can look at it both ways
pKa of H3PO4 = 2.12 so the Ka = 7.58x10^-3
7.58x10^-3 = [H+][H2PO4-] / [H3PO4]
7.58x10^-3 = x^2 / 0.1 - x
x^2 + 7.58x10^-3x - 7.58x10^-4 = 0
x = 0.024M H+
0.024M H+ x 0.15L = 0.0036moles H+
0.4L x 0.2M NaOH = 0.08moles OH-
NaOH neutralizes H+ in a 1:1 ratio therefore, 0.08moles OH- will be neutralized by 0.0036moles H+ leaving 0.0764moles OH-
0.0764moles OH- / 0.55L = 0.139M OH- (the total volume of OH- + H+ = 0.55L)
pH = 14 - pOH
pH = 13.14
the second way includes all of the H+ produced from each of the dissociations of H3PO4
we already have the first dissociation done above.
pKa H2PO4- = 7.21 so Ka = 6.16x10^-8
6.16x10^-8 = [H+][HPO4 2-] / [H2PO4 -]
6.16x10^-8 = x^2 / 0.024 - x.....0.024M from the first dissociation
x^2 + 6.16x10^-8x - 1.48x10^-11 = 0
[H+] = [HPO4 2-] = 3.82x10^-6M
3.82x10^-6M x 0.15L = 5.72x10^-7moles H+ from 2nd dissociation
0.0764moles OH- - 5.72x10^-7moles H+ = 0.0764moles OH-
since the H+ produced by the 2nd dissociation of H3PO4 provides so little H+, the [OH-] doesn't change and the pH will not change, the 3rd dissociation will produce an extremely small amount of H+ and this, too, will not alter the pH
the answer remains 13.14
the other 2 dissociations will produce H+ but it will be negligible to the pH
we can look at it both ways
pKa of H3PO4 = 2.12 so the Ka = 7.58x10^-3
7.58x10^-3 = [H+][H2PO4-] / [H3PO4]
7.58x10^-3 = x^2 / 0.1 - x
x^2 + 7.58x10^-3x - 7.58x10^-4 = 0
x = 0.024M H+
0.024M H+ x 0.15L = 0.0036moles H+
0.4L x 0.2M NaOH = 0.08moles OH-
NaOH neutralizes H+ in a 1:1 ratio therefore, 0.08moles OH- will be neutralized by 0.0036moles H+ leaving 0.0764moles OH-
0.0764moles OH- / 0.55L = 0.139M OH- (the total volume of OH- + H+ = 0.55L)
pH = 14 - pOH
pH = 13.14
the second way includes all of the H+ produced from each of the dissociations of H3PO4
we already have the first dissociation done above.
pKa H2PO4- = 7.21 so Ka = 6.16x10^-8
6.16x10^-8 = [H+][HPO4 2-] / [H2PO4 -]
6.16x10^-8 = x^2 / 0.024 - x.....0.024M from the first dissociation
x^2 + 6.16x10^-8x - 1.48x10^-11 = 0
[H+] = [HPO4 2-] = 3.82x10^-6M
3.82x10^-6M x 0.15L = 5.72x10^-7moles H+ from 2nd dissociation
0.0764moles OH- - 5.72x10^-7moles H+ = 0.0764moles OH-
since the H+ produced by the 2nd dissociation of H3PO4 provides so little H+, the [OH-] doesn't change and the pH will not change, the 3rd dissociation will produce an extremely small amount of H+ and this, too, will not alter the pH
the answer remains 13.14