Question

If you mix 10 mL of a 0.1 M HCl solution with 8 mL of

a0.2 M NaOH solution, what will be the resulting pH?

Answer 1

The pH of the resulting solution is 12.52.

Step-by-step explanation:

`Molarity=(n)/(V)`

n = number of moles

V = volume of the solution in Liters

1)1 mol of HCl gives 1 mol of hydrogen ion.

`[HCl]=[H^+]=0.1 m`

Concentration of the hydrogen ion = 0.1 M

Volume of the solution = 10 mL = 0.010 L

`0.1 M=(n)/(0.010L)`

Moles of hydrogen ions = = 0.001 mol

2) 1 mol of NaOH gives 1 mol of hydroxide ion.

`[NaOH]=[OH^-]=0.2 M m`

Concentration of the Hydroxide ions = 0.2 M

Volume of the solution ,V'= 8 mL = 0.008 L

`0.2=(n')/(V')`

Moles of hydroxide ions ,n ' = 0.0016

1 mol of HCl neutralizes 1 mol of NaOH ,then 0.001 mol of HCl will neutralize 0.001 mol NaOH.

So left over moles of hydroxide ions in the solution will effect the pH of the solution:

Left over moles of hydroxide ions in the solution = 0.0016 mol - 0.0010 mol = 0.0006 mol

Left over concentration of hydroxide ions:

`[OH^-]'=(0.0006 mol)/(0.010 L+0.008 L)=0.0333 mol/L`

`pOH=-\log[OH^-]=-\log[0.03333 M]=1.48`

pH +pOH = 14

pH = 14 - 1.48 = 12.52

The pH of the resulting solution is 12.52.