It is desired to prepare 4.0 M nitric acid from the available acid solution of strength 1.4 M & 6.8 M respectively. If the total volume of the 4.0 M nitric acid required to be prepared is 4 #dm^3#, calculate the volume of the 2 acid solutions to be mixed?

2 Answers
Jul 13, 2015

Answer:

You need #2.074dm^(3)# of the 1.4M solution and #1.926dm^(3)# of the 6.8M solution.

Explanation:

First we find the total moles of #HNO_3# required:

#c=n/v#

So #n=cxxv=4xx4=16mol#

Let volume of the 1.4M solution = #X#

Let volume of the 6.8M solution =#Y#

So we can write:

#16=1.4X+6.8Y#..................#color(red)((1))#

and since the total volume = #4dm^(3)# we can write:

#X+Y=4#...................#color(red)((2))#

These are called simultaneous equations. There are 2 equations and 2 unknowns which we can easily solve:

From #color(red)((2))# we can write:

#X=4-Y#

We can then substitute this for #X# into #color(red)((1))rArr#

#16=1.4(4-Y)+6.8Y#

#Y=10.4/5.4=1.926dm^(3)#

So from #color(red)((2))#:

#X=4-1.926=2.074dm^(3)#

Jul 13, 2015

Answer:

The volumes of the two solutions are 2.1 L and 1.9 L, respectively.

Explanation:

The idea behind this problem is that you need to use two equations, one that describes the number of moles needed for the target solution, and one that describes the total volume of the target solution.

You know that molarity is defined as moles of solute, in your case nitric acid, divided by liters of solution.

The target solution must have a concentration of 4.0 M and a volume of 4 L (#"1 dm"^3 = "1 L"#), so it must contain

#C = n/V => n = C * V#

#n_"total" = 4"moles"/cancel("L") * 4cancel("L") = "16 moles"# #HNO_3#

Now, let's label the volume of the 1.4-M solution #x# and the volume of the 6.8-M solution #y#.

The total volume of the solution must be 4 L, so you know that

#x + y = "4 L"# #" "color(blue)((1))#

Use the molarities and volumes of the stock solutions to find a relationship between the number of moles of nitric acid each solution adds to the total.

#1.4"moles"/cancel("L") * xcancel("L") + 6.8"moles"/cancel("L") * ycancel("L") = "16 moles"#

This is your second equation

#1.4 * x + 6.8 * y = 16# #" "color(blue)((2))#

Use equation #color(blue)((1))# to get a value for #x#

#color(blue)((1)) => x = 4 - y#

Use this value in equation #color(blue)((2))# to get

#1.4 * (4 - x) + 6.8 * y = 16#

#5.6 - 1.4 * y + 6.8 y = 16#

#5.4 * y = 10.4 => y = 10.4/5.4 = "1.926"#

This means that #x# will be

#x = 4 - y = 4 - 1.926 = "2.074"#

Since #x# and #y# represent the volumes of the two stock solutions, the answer will be

#V_"1.4 M" = color(green)("1.9 L")#

#V_"6.8 M" = color(green)("2.1 L")#

I rounded the answer to two sig figs, despite the fact that you only gave one sig fig for the volume of the target solution.