In the reaction between sodium hydroxide and sulphuric acid solutions, what volume of 0.5M sodium hydroxide would exactly neutralise 10cm\(^3\) of 1.25M sulphuric acid?
5cm3
10cm3
20cm3
25cm3
50cm3
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Yes the best formula is CaVaNb= CbVbNa / CbNa we have CaVa Nb/CbNa hence 1.25x10x2/0.5x1= 25/0.5 =50cm3
It'z quite easy. Using the volumetric analysis formulae, CaVa/CbVb =Na/Nb where C, V and N are the concentrations, volumes and moles of the reactants, we get:( 1.25x10)/(0.5Vb) = 1/2. Calculating that will give us 50cm cube. NB: Don't 4get to multiply by 2.
n(a)/n(b)=CaVa/Cb/Vb
2/1=(0.5×Va)/(1.25×10)
Va=2(12.5)/0.5
Va=25/0.5
=50cm3.
QED
Really enjoy u guys,u all try,bt my advice 2 u is 2. use d 2nd formulae 4 fast steppings nd answer
Usin cava/cbvb:na/nb therefore,ca=1.25,va=10cm,cb=0.5,vb=?,na=1 from d equation ,and nb=2 frm d equation therefore,goin to d formular 1.25*10/0.5*vb=1/2 wen u crossmultiply,d ansa u get is 50cm
there's a misunderstanding either from the question or the method of solving or from the equation arrangement.
my: from the question it's NaOH first and H2SO4
Equation: 2NaOH + H2SO4 - NaSO4 + 2H2O
2 : 1
v x 0.5 = 10 x 1.25
2(10 x 1.25) = v0.5
25/0.5 = v; v= 50cm^3 or use CaVa/CbVb = a/b
Since 2 was added to NaOH to balance the equation, in the calculation why is d 2 use to multiplied both volume and mole of the acid??
To derived NA2SO4 as d product, u av to add two @d back of sodium hydroxide(Naoh) to make d eqtn balance..so we derived two as d mole numba of naoh in d frmula of cava/cbvb=na/nb...u gat me rgt.?
The answer is so correct beyond reasonable doubt, to make things more easier use CaVa=Nb/CbVb=Nb formula u will get it right
but the question said sulphuric acid isn't it H2S cos I think H2S is also a sulphuric acid
