Question

What is the h+ ion concentration of a solution prepared by dissolving 4g of naoh in 1000ml?

Answer 1

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student-nameShivam Sharma asked in Chemistry

what will be the H+ ion concentration when 4g NaOh dissolved in 1000 ml water

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student-nameBalaram Shil answered this

in Chemistry, Class

Dear Student,

At first, calculate the [OH-] because [H+] [OH-] = 1 x 10-14.

Given,

amount of solute = 4 g

molar mass of NaOH = 40 g

Vol. of water = 1000 ml =. 1 litre

Now, applying the formula of Molarity for calculating [OH]- , we get

[OH]- = 4/40 x 1

=0.1 mol/litre

Now,

[H]+[OH]- = 1 x 10-14

[H]+ = 1 x 10-14 / 0.1

=1 x 10-13 M

Hope this information will clear your doubts about topic.

Answer 2

Given:

Amount of NaOH = 4g

Volume of water = 1000ml = 1 litre

To find:

$H^{+}$ ion concentration of a solution prepared by dissolving 4g of NaOH in 1000ml of water.

Solution:

We know that, $\text{Molarity}&=\frac{\text{Number of moles of the element}}{\text{Volume of solution in litres}}$

Number of moles of Na in NaOH = $\frac{\text{Mass in grams}}{\text{Molar mass}}$ = $\frac{4}{40} = 0.1 mole.$

Molarity = [$OH^{-}$] =  $\frac{0.1}{1} = 0.1 mol/litre.$

So, the molarity is 0.1 or $10^{-1}$ which is equal to the concentration of hydroxyl ions.

We know that,

$\begin{array}{l}{\left[H^{+}\right]\left[O H^{-}\right]=10^{-14}} \\\\{\left[H^{+}\right]=\frac{10^{-14}}{\left[O H^{-}\right]}} \\\\{\left[H^{+}\right]=\frac{10^{-14}}{10^{-1}}=10^{-13}}\end{array}$

Therefore, the concentration of $H^{+}$ ions is $10^{-13}$.