Question

Find [H+] ion concentration in 100 ml of 0.001M NaoH solution

Answer 1

Answer : The concentration of hydrogen ion is, $1\times 10^{-10}M$

Solution :

First we have to calculate the moles of NaOH.

$\text{Moles of NaOH}=\text{Molarity of NaOH}\times \text{Volume of NaOH}=0.001mole/L\times 0.1L=1\times 10^{-4}moles$

As we know that NaOH dissociates into $Na^+$ ions and $OH^-$ ions.

As, 1 mole of NaOH dissociates 1 mole of $OH^-$ ions.

So, $1\times 10^{-4}moles$ mole of NaOH dissociates $1\times 10^{-4}moles$ mole of $OH^-$ ions.

Now we have to calculate the pOH.

$pOH=-\log [OH^-]\\\\pOH=\log (1\times 10^{-4})\\\\pOH=4$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4=10$

Now we have to calculate the $H^+$ ions concentration.

$pH=-\log [H^+]\\\\10=-\log [H^+]$

$[H^+]=1\times 10^{-10}M$

Therefore, the concentration of hydrogen ion is, $1\times 10^{-10}M$

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.