What is the normality of 0.15 M NaOH?

Relation between Normality and Molarity is:

N=n(M)

Where n is acidity, or basicity

Acidity of NaOH=1

N=1(0.15)

N=0.15

1)For NaOH :

NaOH is a monoacidic base as 1 mole of it reacts with only one mole of H(+) ions for complete neutralization.

NaOH + H(+) --> Na(+) + H2O

So, equivalent mass of NaOH is same as its molar mass.

Normality of NaOH = Molarity = 0.1381 N

2) For H3PO4 :

H3PO4 is a tribasic acid as 1 mole this acid can give 3 moles of H(+) ions on complete ionisation.

Hence, equivalent mass of H3PO4 = Molar mass/3

Normality of H3PO4 = Molarity x 3 = 0.0521 x 3

= 0.1563 N (answer)

In chemistry,

We denote Molality with m and Molarity with M.

Where, Molality(m)=No. of moles of solute/Amount of solvent (in kg)

And Molarity(M)=No. of moles of soulte/Volume of solution(in Litres)

Therefore,

0.5m of NaOH means 0.5 moles of NaOH in 1 kg of solvent

And, 0.5 M of NaOH means 0.5 moles of NaOH in 1 Litre of Solution

Formula: M=n/V

Given:

M: 0.3

V: 0.3L

We need to find n and then convert into grams.

So,

n=M*V

n= (0.3)(0.3)

=0.09 mol

Now the molar mass of NaOH is 39.997 g/mol so use that to convert mols into grams.

0.09 mol * 39.997g/mol = 3.6 g

So measure out 3.6g of NaOH and add water to make 300mL of solution!

Since sodium hydroxide is a strong base, it tends to completely dissociate in aqueous solution. The formula NaOH has only one ion of Na+ and one of OH-. So the multiplier to figure out the Normality is also 1.

See https://www.thoughtco.com/definition-of-normality-in-chemistry-605419

M1 = Molarity of initial solution = 15 M,

V1 = Volume of initial solution = 10 mL = (10/1000) L,

M2 = Molarity of resulting solution = ?,

V2 = Volume of resulting solution = 0.50 L.

Dilution Formula: M1V1 = M2V2

(10/1000) x 15 = M2 x 0.50

M2 = (10/1000) x 15/0.50 = 0.30

So, the concentration of resulting solution = 0.30 M

Hope, this helps.

0.15 M NaOH is 0.15 Normal in Na+ and 0.15 Normal in OH-.