Question

Find the [H+] ion concentration in 100mL of 0.001 M NaOH solution.

Answer 1

number of moles = volume of solution × concentration

given, volume of solution = 100mL

concentration = 0.001M

so, number of moles = 100 × 0.001 = 0.1 mili mole

molarity or concentration in 1L = number of moles/1000mL

= 0.1/1000mL

= 10^-4 M NaOH solution

$NaOH\Leftrightarrow Na^+ +OH^-$

so, pOH = -log[OH^-]

= -log(10^-4) = 4

we know, pH + pOH = 14

so, pH = 14 - pOH = 14 - 4 = 10

or, pH = -log[H^+] = 10

or, [H^+] = 10^-10

hence, concentration of [H^+] ion is 10^-10 M

Answer 2

Answer:

Explanation:

Quantity of solution = 100ml (Given)

Concentration = 0.001ml (Given)

Number of moles = volume of solution × concentration

= 100 × 0.001 = 0.1 mili mole  

Molarity or concentration in 1L = number of moles/1000mL  

= 0.1/1000  

= 10~-4 M NaOH solution  

Thus, pOH = -log[OH^-]  

= -log(10~-4)  

= 4  

Since, pH + pOH = 14  

Thus, pH = 14 - pOH

= 14 - 4  = 10

or, pH = -log[H~+] = 10

or, [H~+] = 10^-10  

Therefore, the concentration of [H~+] ion will be 10~-10 M