February 20, 2017

Homework Help: Chemistry

Posted by Mike on Monday, April 11, 2011 at 8:40pm.

If a researcher needs 0.50 L of 3.0 M nitrite buffer at pH 3.00, then she knows that the concentration of nitrous acid (conjugate acid) plus the concentration of nitrite anion (conjugate base) must equal 3.0 M. She must then calculate the amount of each that is necessary to give this total amount and, at the same time, yield a pH = 3.00. Using the Henderson-Hasselbach equation:
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?

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