Complete the following table by calculating the missing entries and indicating whether the solution is, acidic or basic. (Make sure you show your work!) OH1 Acidic or basic? [H\'1 pH pOH 6.8 x 10-3 M 2.1 x 109 M 6.73 4.25 HTML Editor OH-] Acidic or Basic? pH pOH 6.8 x 103 M 2.167 2.1 x 109 M 6.73 4.25 table tbody tr tod 19 words Solution [H+] = 6.8x10^-3M [H+][OH-] = Kw           where Kw = ionic product of water = 1.0x10^-14 [OH-] = Kw/[H+] = 1.0x10^-14/6.8x10^-3 = 1.47x10^-12 [OH] = 1.47x10^-12M [H+] = 6.8x10^-3 -log[H+] = -log( 6.8x10^-3) PH= 2.167 PH+POH= 14 POH= 14-PH POH= 14-2.167 =11.833 [H+] = 6.8x10^-3M,   [OH-]= 1.47x10^-12M, PH= 2.167 ,       POH= 11.833 PH<7, so the nature of solution is acidic. b) [OH-] = 2.1x10^-9M [H+] = 1.0x10^-14/2.1x10^-9 = 4.76x10^-6 [H+] = 4.76x10^-6M -log[H+] = -log( 4.76x10^-6) PH= 5.322 POH= 14-5.322 POH= 8.678 [H+] = 4.76x10^-6M , [OH-] = 2.1x10^-9M ,         PH= 5.322,             POH= 8.678 PH<7, so the nature of the solution is acidic c) PH= 6.73 -log[H+] = 6.73 [H+] = 10^-6.73 [H+] = 1.86x10^-7M [OH-] = 1.0x10^-14/1.86x10^-7 [OH-] = 5.376x10^-8M POH= 14-6.73 POH= 7.27 [H+] = 1.86x10^-7M.   [OH-] = 5.376x10^-8, PH= 6.73,   POH= 7.27 PH<7, so the nature of the solution is acidic nature, d) POH=4.25 PH= 14-4.25=9.75 PH= 9.75 -log[H+] = 9.75 [H+] = 10^-9.75 [H+] = 1.778x10^-10M [OH-] = 1.0x10^-14/1.778x10^-10 [OH-] = 5.62x10^-5 [H+] 1.778x10^-10M, [OH-] = 5.62x10^-5M , PH= 9.75, POH= 4.25 PH>7, so the nature of the solution is basic . .

1. Complete the following table by calculating the missing entries and indicating whether the
solution is, acidic or basic. (Make sure you show your work!) OH1 Acidic or basic? [H'1 pH
pOH 6.8 x 10-3 M 2.1 x 109 M 6.73 4.25 HTML Editor OH-] Acidic or Basic? pH pOH 6.8 x
103 M 2.167 2.1 x 109 M 6.73 4.25 table tbody tr tod 19 words
Solution
[H+] = 6.8x10^-3M
[H+][OH-] = Kw where Kw = ionic product of water = 1.0x10^-14
[OH-] = Kw/[H+] = 1.0x10^-14/6.8x10^-3 = 1.47x10^-12
[OH] = 1.47x10^-12M
[H+] = 6.8x10^-3
-log[H+] = -log( 6.8x10^-3)
PH= 2.167
PH+POH= 14
POH= 14-PH
POH= 14-2.167 =11.833
[H+] = 6.8x10^-3M, [OH-]= 1.47x10^-12M, PH= 2.167 , POH= 11.833
PH<7, so the nature of solution is acidic.
b) [OH-] = 2.1x10^-9M
[H+] = 1.0x10^-14/2.1x10^-9 = 4.76x10^-6
[H+] = 4.76x10^-6M

2. -log[H+] = -log( 4.76x10^-6)
PH= 5.322
POH= 14-5.322
POH= 8.678
[H+] = 4.76x10^-6M , [OH-] = 2.1x10^-9M , PH= 5.322, POH= 8.678
PH<7, so the nature of the solution is acidic
c)
PH= 6.73
-log[H+] = 6.73
[H+] = 10^-6.73
[H+] = 1.86x10^-7M
[OH-] = 1.0x10^-14/1.86x10^-7
[OH-] = 5.376x10^-8M
POH= 14-6.73
POH= 7.27
[H+] = 1.86x10^-7M. [OH-] = 5.376x10^-8, PH= 6.73, POH= 7.27
PH<7, so the nature of the solution is acidic nature,
d)
POH=4.25
PH= 14-4.25=9.75
PH= 9.75
-log[H+] = 9.75
[H+] = 10^-9.75

3. [H+] = 1.778x10^-10M
[OH-] = 1.0x10^-14/1.778x10^-10
[OH-] = 5.62x10^-5
[H+] 1.778x10^-10M, [OH-] = 5.62x10^-5M , PH= 9.75, POH= 4.25
PH>7, so the nature of the solution is basic .