Question

Complete the following table: 1. Acidic or Basic [H*]. M IOн, М pH РОН 2.45 5.82 x 10-6 -0.11 5.4 x 10 3.25 -206 5.78 x 10-10 2.26 4.93 x 10 7.00

Can you explain how to do this please

Answer 1

Ofcourse,I will explain.it is so simple.

First you need to know these equations to solve this okay.

1. pH + pOH = 14

2.pH = -log [H⁺]

3.pOH = -log [ OH^{​​​​​-}​]

by using above 3 formula you can do that easily.

I will do it for one example from each of the column so you get the whole thing okay.I will take the example which is harder one.

- О.206. pH= 2. — о. 1 Рон S0, 30, : 19 PH рон pH+ pОн 3 1Ч Рон- 14-рH 14-(-0.206 I PH O. = 14 PHЕ 14. 0 - 14 +0.206 Рон- 14.206 МOW, NOCJ, PH -log CH*J PH= - log CHJ. : -0.206 : -1og СHt] 14. -log CHtJ log CH J -14.11 log CHJ .. СнJ : amtilog roao6) THJ.61 m .. о.206 - :. CHJ = an ti log (-14.11) . СHТJE 7. 76 x10 Cund, cmd, pон: -lоg Cон] Рон - 1оg Coн) -о. м -lоg Cон] 5. 14.206 : -l0g Cон ] о.1Е log Сон ] log Coн ] :. - 14.206 :. Сон"]: аmtilog ra) . Сон): amti logc-14.200) . Сон ]- 6.22 х 10'5 .. Сон ] 3 1.2882 m неsе, рн& + Неъе, рн >7 So, t is ucidic basic So, it is

3. СонJ Чo -IO CHJ S.T8 x1o log CS.4 x10J PH-log CHJ -loq Cs.78 xio POH 9logio-log S. 9-O.7323 POH 8. 27 10 - O.1619 pH 9. 24 е. рн +рон -14 . pH 14- pOH pH 14 9.2 PH S.3 POH 4.16 . pH -log THJ 5. рон г -log Coн] .4.16-loq COHJ . S.13 log CH J 4.16 og CoH) 1.3103 CH* J CHJati log ES.73) .COHJ n tilog C-4.16) . -6 CHJ 6 x1o COHJ.74x10 неъе, рн > 7 Hee PH <7 is basic s0, it is cidic

Do this for all the other quantity , you will get the answers .

for the last column , see the pH values

if pH > 7 then basic,

if pH <7 then acidic,

if pH= 7 then neutral(last row).

PLEASE GIVE A THUMBS UP IF IT REALLY HELPS YOU.

THANK YOU.