Using the correct amount of significant figures, how many milliliters of 8.48×10^(−2) M Ba(OH)2 (aq) are required to titrate 54.90 mL of 5.38×10^(−2) M H(NO)3?

1 Answer
Jul 14, 2016

Answer:

#sf(17.4color(white)(x)ml)#

Explanation:

Start with the equation:

#sf(Ba(OH)_2(aq)+2HNO_3(aq)rarrBa(NO_3)_2(aq)+2H_2O(l))#

So 1 mole of #sf(Ba(OH)_2)# will react with 2 moles of #sf(HNO_3)#

Concentration = amount of solute / volume of solution.

So:

#sf(c=n/v)#

#:.##sf(n=cxxv)#

#:.##sf(nHNO_3=5.38xx10^(-2)xx54.90/1000=2.953xx10^(-3))#

#:.##sf(nBa(OH)_2=(2.953xx10^(-3))/2=1.477xx10^(-3))#

#sf(v=n/c=(1.477xx10^(-3))/(8.48xx10^(-2))=0.01742" "L)#

#sf(v=17.4color(white)(x)ml" ""to 3sf")#.