jkn1121

A. I think the pH range as of 9.6 from 9-10 for the buffer since the buffer keeps the system at equilibrium and the pH doesnt change that much when the buffer is added it. B. If I piggyback on what you did, the concetration is .2511 [-NH3] then it should be 75% NH3, 25% NH2 I don't unerstand how it went from ([g-NH2]/[g-NH3+]) to ([g-NH3]/[g-NH3+])

A man used a hydrogen-filled balloon to fly from Paris 25km into the French side. What’s the mass density of hydrogen relative to air at the same temperature/pressure? What mass of payload can be lifted by 10kg of hydrogen, neglect the mass of the balloon?

So you are saying in A. ranges from pH is 10.6 and 8.6. As for B pH = pKa + log ([A-]/[HA]) 9= 2.51x 10^-10 + log (A/HA)?

I still don't understand what you mean

You have to find the concentration of A, HA, in order to solve fo pH if the equation is pH = pKa + log ([A-]/[HA]). So for Part A the solution is quite basic. It can range from 7-14. Then using the 0.1M in pH= 9.0 thats consider the A in the formula, so you plug it in, to get 9= 9.6 + log (0.1/HA)?

yea, i know that pka 9.6 is like the pH is 2.511x 10^-10 which quite small and the pka is basic so the the pOH is 4.4 which is 3.98x 10^-5.

The aminoacid glycine is often used as the main ingredient of a buffer in biochemical experiments. The amino group of glycine, which has a pKa of 9.6, can exist either in the protonated form ( -NH3+) or as the free base (-NH2), because of the reversible equilibrium R-NH3+ Û R-NH2 + H+ a) In what pH range can glycine be used as an effective buffer due to its amino group B) In a 0.1 M solution of glycine at pH 9.0, what percentage of glycine has it’s amino group in the NH3+ form? c) How much of 5 M NaOH must be added to 1 L of 0.1 M glycine at pH 9.0 to bring its pH to exactly 10.0? d) When 99% of the glycine is in its –NH3+ form, what is the numerical relation between the pH of the solution and the pKa of the amino group? (for example, “the pH is double of the pKa” or “the pH is 1 pH unit above the pKa”, or “pH = pKa + 1”, etc.) I dont know where to begin.