How do I make a 1N H2SO4 solution?

(.1N) H2SO4 ----- 2x 0.1 = 0.2N H+ ion

pH = -log[H+], = -log 2 x 10^-1 = 1 —log2 = 0.6990.

by mass by volume??

from stock????

D = 1.84 g/mL 98%

diluted 5% D = 1.035 g/mL

%C = m / DV (100)

1.84 x V = 100

V = 54.35 mL concnetrated

%C1V1D1 = %C2V2D2

0.98 x 1,84 x 54.35 = 0.05 x 1.05 (V2 + 54.35)

V2 = 1,812.3 mL

take 54.35 mL of the concentrated acid and dilute (acid to water) to final cvolume of 1,866.7 mL

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1. H2SO4(aq) = 2H+ + SO4- -
2. From eq #1 1mole of H2SO4 provides 2 moles of H+
3. So 98 g/mole of H2SO4 = 2M of H+ or 2eq
4. 1/2 x 2M = 1N (one eq)
5. 1/2 x 98g.mole = 49g of H2SO4 diluted to 1L with water
6. Add 49g + ((1/%H2SO4) x 100) and add slowly with stirring to ~500ml of water.
7. Cool. Then add water to 1000ml mark.
8. Shake

a) V1N1=V2N2

V1(36.8)= V2(2.5)

V2 = 14.72V1

Dilute a desired volume from the initial solution by a factor of 14.72 (acid to water)

b) normality definition

X/49/1L = 2.5

X = 122.5 g H2SO4

from the initial solutionn we need 122.5 g

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b)

D=1.84

%m/m = 98

m=%CDV

122.5 = 0.98 x 1.84V

V = 67.94 mL

take this volume from the initial solution and dilute till final volume of 1 L (acid to water)

Because of the H2 bit of the equation.

A 1 molar solution is the molecule mass of the solute (in this case sulphuric acid ) in 1 litre of water.

A 1 N solution (with respect of H) contains 1 mole of H ions in 1 litre of water. In the case of sulphuric acid, that would be 0.5 moles of sulphuric acid in 1 litre of water, because the formula for sulphuric acid is H2SO4.

1 mole of Sulphuric acid in 1 litre of water would be 1M, and 2N with respect to H ions, but 1N with respect to SO4 ions. (We tend to imagine that N is always with respect to H so the matter gets a little fudgy).

$1•N$ sulfuric acid represents $0.5•mol•L^{-1}$ sulfuric acid. Why? Because $ext{normality}$ refers to the number of equivs of base required to reach neutrality … and thus sulfuric acid reacts with 2 equiv of say sodium hydroxide to give a neutral solution…

$H_{2}SO_{4}(aq) + 2NaOH(aq) longrightarrow Na_{2}SO_{4}(aq) + 2H_{2}O(l)$

And so to make a LITRE volume of $0.5•mol•L^{-1}$, we take $27•mL$ of conc. sulfuric (i.e. $18•mol•L^{-1}$), and add this to $973•mL$ of distilled water ... and the order of addition, acid TO water, is VITALLY important...

from concentrated???

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VM = # mol

0.05 x 3 = 0.15 mol wee need

0.15 = 18.4V

V = 8.15 mL

take this volume from thecoentrated acid and dilute till 50 mL(acid to water)

1N H2SO4 means 1 equivalent of H2SO4 is present in 1 litre of solution …

Equivalent weight of H2SO4 = 49

Means 49 gm of H2SO4 is 1 equivalent of H2SO4 .

So just put 1 litre of pure water and mix it carefully with 49 gm of H2SO4, now you have 1N H2SO4 solution ….

This problem also puzzles me. I am struggling with my own question

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What is the concentration of the HCl solution.

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Seminormal solution is formed by dissolution of 2 gram equivalent of acid in 1 litre of water. Gram equivalent of acid =( weight of acid *its basicity (2) )/molecular weight. In that case, 2*98/2 will be the weight of the acid.

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