Question

Food [H3O+] [OH-] pH Acidic,Basic or Neutral

Asparagus 4.8x10-6M

Tomatoes 4.23

Milk 5.4 x10-7M

Complete the following table for a selection of foods: Calculaye the [H3O+] of asparagus’ with the [OH] =4.8x10-7M ?

Calculate the Ph of asparagus with the [OH-]=4.8x10-6M?

Calculate the [H3O+] of tomatoes with the pH=4.22

Calculate the [OH-] of tomatoes with the ph=4.22

Calculate the OH- of the milk with the H3O+=5.4x10-7M

Calculate the ph of the milk with the H3O+=5.4x10-7M

Answer 1

1)

we have below equation to be used:

[H3O+] = Kw/[OH-]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[H3O+] = (1.0*10^-14)/[OH-]

[H3O+] = (1.0*10^-14)/4.8E-7

[H3O+] = 2.08*10^-8 M

2)

we have below equation to be used:

pH = -log [H3O+]

= -log (4.8*10^-6)

= 5.32

3)

we have below equation to be used:

pH = -log [H3O+]

4.22 = -log [H3O+]

log [H3O+] = -4.22

[H3O+] = 10^(-4.22)

[H3O+] = 6.03*10^-5 M

4)

we have below equation to be used:

pH = -log [H+]

4.22 = -log [H+]

log [H+] = -4.22

[H+] = 10^(-4.22)

[H+] = 6.026*10^-5 M

we have below equation to be used:

[OH-] = Kw/[H+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H+]

[OH-] = (1.0*10^-14)/(6.026*10^-5)

[OH-] = 1.66*10^-10 M

5)

we have below equation to be used:

[OH-] = Kw/[H3O+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H3O+]

[OH-] = (1.0*10^-14)/(5.4*10^-7)

[OH-] = 1.85*10^-8 M

6)

we have below equation to be used:

pH = -log [H3O+]

= -log (5.4*10^-7)

= 6.27