I’m trying to find the molarity of h2so4 in this calculation. i attempted it but i still need someone to check my work. i used 10.00 ml of h2so4 and the volume added to it was 24.97 ml of the 0.0924m naoh solution. the equation is 2naoh + h2so4 -> 2h2o + na2so4, so i’m aware 1 mol of h2so4 is 2 mol naoh. what i tried: 0.0924 mol/l x 0.02497 l= 0.002307 mol naoh x 1 mol h2so4/2 mol naoh= 0.001156 mol h2so4 0.0011536 mol h2so4/0.01000 l h2so4= 0.1154m h2so4 am i doing this right?

Since it goes to completion, we can use:  

Use NaMaVa =NbMbVb  

where  

Na= moles of H+ in H2SO4  

Ma= molarity of the acid  

Va= volume of the acid  

Nb= moles of OH- in NaOH  

Mb= molarity of the base  

Vb= molarity of the base  

plug in the numbers  

(2 moles of H+) (Volume acid) (10.00mL acid) = (1 mole of OH-) (.248M base) (18.71mL base)  

20 V = 4.64  

V= .232 M

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%