Question

Need to complete the chart

Answer 1

Row 1

`[H^+]=1.8* 10^(-6)M`

`pH=-\log[H^+]=-\log[1.8* 10^(-6)]=5.7`

pOh=14-pH=14-5.7=8.3

`pOH=-\log[OH^-]`

`[OH^-]=0.5* 10^(-8)M`

Hence, acidic

Row 2

`[OH^-]=3.6* 10^(-10)M`

`pOH=-\log[OH^-]=-\log[3.6* 10^(-10)]=9.4`

pH=14-pOH=14 - 9.4 = 4.6

`pH=-\log[H^+]`

`[H^+]=2.6* 10^(-5)M`

Hence, acidic

Row 3

pH = 8.15

`[H^+]=0.7* 10^(-8)M`

pOH=14-pH=14 - 8.15 = 5.8

`pOH=-\log[OH^-]`

`[OH^-]=1.5* 10^(-6)M`

Hence, basic

Row 4

pOH = 5.70

`[OH^-]=1.8* 10^(-6)M`

pH=14-pOH=14 - 5.70 = 8.3

`pH=-\log[H^+]`

`[H^+]=0.5* 10^(-8)M`

Hence, basic