Chemistry, 16.06.2020 04:57 zay65307

What volume of 0.110 M Hydrochloric acid needs to be added to 30 mL of 0.100 M ammonia to make a buffer with a pH of 9.30? The Kb value for NH3 is 1.8 x 10^-5 .

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Answer from: almostnevercbbc

12.937 mL

Explanation:

First, we have to start with the hendersson-hasselbach equation:

$pH=pKa+\frac{A^-}{HA}$ (Equation 1)

So, from this equation, we already know the desired pH value 9.3. Therefore we have to calculate the pKa value, for this, we need to use the following equations:

pKb=-LogKb (equation 2) and 14=pKa+pKb (equation 3) With this in mind we can do the calculations:

pKb=-Log(1.8x10^-^5)=4.74

pKa=14-4.74=9.26

Additionally, we have to know the buffer system reaction:

NH_4^+~~NH_3~+~H^+ (reaction 1)

With this in mind NH_4^-=HA and NH_3=A^- . Now if we use the hendersson-hasselbach reaction we can find the ratio ( $\frac{A^-}{HA}$ ) that we need for the desired pH value, so:

$9.30=9.26+Log(\frac{A^}{HA})$

$\frac{A^-}{HA}=1.108$ (equation 4)

From the problem, we know that we already have a portion of NH_3=A^- , 30 mL and 0.1 M. Also, the HCl that we are going to add will react with the NH3 ( H^+=HCl ), so:

H^+~+~NH_3=~NH_4^+ (reaction 2)

With this in mind, all the HA concentration will come from the addition of the HCl and this addition will consume some A-. Therefore, we have to know the moles of A- before the addition:

moles~of~A^-~=~0.03*0.1=0.003~mol~of~A-

So, we can write a new equation that calculates the concentration of A^- before the addition:

$A^-=\frac{0.003-X}{0.03+Y}$ (equation 5)

In this case, "X" is the of A- (NH3) consumed in the reaction for the addition of HCl and Y is the volume of HCl (in litters). Additionally, we can write another equation for the concentration of HA after the addition of HCl:

$HA=\frac{X}{Y+0.03}$ (equation 6)

Let's remember that we already have 0.03L (30mL) on the beaker, thats why we have to add "0.03" in the bottom. Now, we can include these last two equations on equation 4, so:

$\frac{\frac{0.003-X}{0.03+Y}}{\frac{X}{Y+0.03}}=1.108$ (equation 7)

We have 2 unknows and 1 equation we need another equation to solve this. So, all X would be produced by the addition of HCl therefore "X" are the moles of HCl added and Y would be the volume of HCl, if we have a concentration of 0.11M for HCl we can use the molarity equation to relate "X" and "Y", so:

$0.11M=\frac{X}{Y}$ (equation 8)

X=0.11*Y (equation 9)

Now, we can replace equation 9 in equation 7, so:

$\frac{\frac{0.003-0.11*Y}{0.03+Y}}{\frac{0.11*Y}{Y+0.03}}=1.108$ (equation 10).

If we do some math we will obtain:

$\frac{0.003-0.11*Y}{0.11*Y}=1.108$ (equation 11)

When we solve for "Y" we obtain a value of 0.012937 L or 12.937 mL.

I hope it helps!

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What volume of 0.110 M Hydrochloric acid needs to be added to 30 mL of 0.100 M ammonia to make a buf...