Question

please answer the following questions, thank you

Pre-lab Calculations 1. What volumes of 1.0 M HNO, and 1.0 M triethanolamine are required to prepare 1.0 L of buffer that has a pH of 7.76 and is 0.020 M in both triethanolamine and triethanolammonium ion? [A] + [HA] = 0.040 mol/L (40 mM) 2. What weight of NaCl must be weighed out to prepare 250 mL of 0.10 M NACI?!

Answer 1

Answer of question no. -1:

It is stated that, the solution must contain 0.02 M triethanolamine and 0.02 M triethanolammonium ion in total 1000 ml volume.

Of course, triethanolammonium ion could be generated from the reaction between HNO₃ and triethanolamine.

HNO₃ + N(CH₂CH₂OH)₃ = HN⁺(CH₂CH₂OH)₃NO₃^-

So, as per the above equation, 1 M HNO₃ will quench 1 M triethanolamine.

Therefore, in order to get a mixture of 0.02 M triethanoammonium ion and 0.02 M triethanolamine with a total volume of 1 L, one must mix 500 ml 0.04 M HNO₃ with 500 ml 0.08 M triethanolamine; so that 500 ml 0.04 M HNO₃ can react with 250 ml 0.08 M triethanolamine along with 250 ml 0.08 M unreacted triethanolamine.

Of course, 250 ml 0.08 M triethanolamine = 0.02 M triethanolamine in 1000 ml solution

500 ml 0.04 M HNO₃ = 0.02 M HNO₃ in 1000 ml solutionand 500 ml 0.08 M trirthanolamine = 0.02 M triethanolamine in 1000 ml solution

So as per the formula, V₁S₁ = V₂S_2, for HNO₃ we have,

500 ml x 0.04 M = V₂ x 1 M

or, V₂ = 20 ml HNO₃

Again for triethanolamine we have,

500 ml x 0.08 M = V₂ x 1 M

or, V₂ = 40 ml triethanolamine

So, we need to mix 20 ml 1 M HNO₃ with 40 ml 1 M triethanolamine in 940 ml water, in order to get 1 L buffer.

Answer of question no. - 2:

250 ml 0.1 M NaCl = 1000 ml 0.025 M NaCl

Now, 1000 ml 1 M NaCl contains 1 gm mole NaCl, i.e. 58.44 gm NaCl

Therefore, 1000 ml 0.025 M NaCl contains (0.025 x 58.44) gm = 1.461 gm NaCl

So, 1.461 gm NaCl must be weighed out to prepare 250 ml 0f 0.1 M NaCl.