A) calculate the ph of a 1.8 * 10 -4 m hcl solution b) calculate the [h +] and poh of a 0.05 m koh solution

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A) 3.745

B) 2 x 10^-13 M [H+]

pOH = 1.3

Explanation:-

A)From the question we see

The strength of HCl solution = 1.8 x 10^-4 M.

The M stands for molarity which is moles / Litre.

HCl is a monobasic acid. So the number of moles of Hydrogen ion H+ HCl can give is the same as the value of it’s strength in moles per litre.

The hydrogen ion concentration [H+] is therefore 1.8 x 10^-4.

The formula for pH is

pH = - log [ H+ ]

= - log [1.8 x 10^-4]

= 3.745

B)From the question we see

The strength of KOH = 0.05 M

The M stands for molarity which is moles / Litre.

KOH is a monoacidic base. So the number of moles of Hydroxyl ion OH- KOH can give is the same as the value of it’s strength in moles per litre.

The Hydroxyl ion concentration [OH-] is therefore 0.05

The relation between [OH-] and [H+] is

[H+] x [ OH-] = 10^-14

[H+] = 10^-14 / [OH-]

= 10^-14/ 0.05

= 2 x 10^-13 M

Using the relation between pH and [OH-]

pOH = - log [OH-]

= -log [0.05]

= 1.3

c is the answer because when something is left after something evaporates that shows the chemical change.