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February 7, 2016
Posted by **Mike** on Monday, April 11, 2011 at 8:40pm.

pH = pKa + log ((base)/(acid))

3.00 = 3.39 + log (NO2-/HNO2)

since NO2- = 3.0 – HNO2

3.00 = 3.39 + log ((3-HNO2)/HNO2)

solving for HNO2 you get 2.13 M, therefore NO2- = 3 – 2.13 = .87 M

I understand all this except where does the 3.39 come from? I assume it is pKa?

Sodium nitrite is a solid, so she would calculte that she needs 30.0 grams to be dissolved in 0.5 L of solution. How does she calculate the 30.0 grams??

If, for example, a solution of 10.0 M HNO2 were in the lab, she would need to use 106 ml of that solution dissolved in the same 0.5 L as the 30.0 g of NaNO2. Where do they get the 106 ml from?

Suppose she wanted to change the pH of the nitrite buffer from pH 3.00 to 2.80, an acid would have to be added to the buffer. What volume of 12 M HCL is needed to make this change? From the Hasselbah equation we know 2.80 = 3.39 + log ((.87-x)/(2.13 +x)) so x =.26, she calculates that whe would need about 11 ml of 12 M HCL. I think I got this because 2.8/.26 almost equals 11, right??

Final question: Above we found that by adding 11.0 mLof 12.0 M HCL the pH of the nitrite buffer was changed from 3.00 to 2.80. Final question: How many grams of solid sodium hydroxide would the chemist add to the buffer to change the pH from 3.00 to 3.80? Below is what I’m thinking but I could be all wrong:

3.80 = 3.00 + log ((3.80-HNO2)/HNO2

solving for HNO2 you get .52 M, therefore NO2- = 3.80 –.52 =3.28 M

How many grams?

- Chemistry -
**DrBob222**, Monday, April 11, 2011 at 10:32pmI understand all this except where does the 3.39 come from? I assume it is pKa?

**Yes, 3.39 is the pKa**

Sodium nitrite is a solid, so she would calculate that she needs 30.0 grams to be dissolved in 0.5 L of solution. How does she calculate the 30.0 grams??

**0.87M x 0.5L = moles NaNO2.**

moles NaNO2 x molar mass NaNO2 = 30 g.

If, for example, a solution of 10.0 M HNO2 were in the lab, she would need to use 106 ml of that solution dissolved in the same 0.5 L as the 30.0 g of NaNO2. Where do they get the 106 ml from?

**How many moles HNO2 do you need? M x L = 2.13 x 0.5L = 1.065 moles. Then M = moles/L and rearrange to L = moles/M = 1.065/10 = 0.1065 which rounds to 106 mL.**

Suppose she wanted to change the pH of the nitrite buffer from pH 3.00 to 2.80, an acid would have to be added to the buffer. What volume of 12 M HCL is needed to make this change? From the Hasselbah equation we know 2.80 = 3.39 + log ((.87-x)/(2.13 +x)) so x =.26, she calculates that whe would need about 11 ml of 12 M HCL. I think I got this because 2.8/.26 almost equals 11, right??

**0.26M x 0.5L = 0.13 moles HCl needed.**

M HNO3 = moles/L or L = moles/M = 0.13/12M = 0.0108L or 10.8 mL which rounds to 11 mL.

Final question: Above we found that by adding 11.0 mLof 12.0 M HCL the pH of the nitrite buffer was changed from 3.00 to 2.80. Final question: How many grams of solid sodium hydroxide would the chemist add to the buffer to change the pH from 3.00 to 3.80? Below is what I’m thinking but I could be all wrong:

3.80 = 3.00 + log ((3.80-HNO2)/HNO2

solving for HNO2 you get .52 M, therefore NO2- = 3.80 –.52 =3.28 M

How many grams?

**I wouldn't do it that way and I'll show you why. First, I went through your calcn and the math part seems to be OK; however, you aren't looking for HNO2 or nitrite. You are looking for NaOH. If we substitute your numbers we get pH = 3.39 + log(3.28/0.52) = 4.19 and you wanted 3.80 so those numbers won't get it. After a lot of looking I finally saw what is going wrong. First, you substituted 3.00 for pKa instead of 3.39. The numbers you obtained are correct for that substitution. The second error you made is substituting 3.80-x for HNO2; it should be 3.00-x for HNO2. If you substitute correctly as 3.80 = 3.39 + log(30-x)/(x) and solve for x you obtain 0.84 for HNO2 and 2.16 for NO2^-. But you CAN do it this way. At this point you must realize that the final HNO2 must be 0.84 so you add 2.13-0.84 = 1.29 = (OH). For NO2- it is 3.00-0.84 = 2.16 total which means you add 1.29 M OH.**

I would do it this way:

...........HNO2 + OH^- ==>NO2^- + H2O

initial....2.13....0......0.87.......

add.................x................

change......-x.....-x......+x.......+x

equil....2.13-x .....0.....0.87+x.....

3.80 = 3.39 + log(0.87+x/2.13-x)

x = 1.29M = OH^-

1.29M x 0.5L x 40(the molar mas) = ??g.

This way the answer comes out directly in M OH to add instead of calculating HNO2 and NO2 and then subtracting from the original value to obtain the amount that must be added.

- Chemistry -
**Mike**, Tuesday, April 12, 2011 at 7:29amThank you so much! I am still very shaky on this, but I understand a lot better!