All Questions
7
questions
0
votes
0
answers
65
views
How to find pH for acetate buffer?
Calculate the pH of the solution that results when $\pu{40.0 mL}$ of $\pu{0.100 M}$ $\ce{CH3COOH}$ is mixed with $\pu{30.0 mL}$ of $\pu{0.200 M}$ $\ce{NaCH3COO}.$ $K_\mathrm{a}(\ce{CH3COOH}) = \pu{1....
- 27
1
vote
0
answers
87
views
How do two species from the same buffer system have different pKa values? [closed]
$$
\begin{array}{lr}
\hline
\text{Reagent}~(\pu{0.1 M}~\text{solutions}) & \mathrm{p}K_\mathrm{a} \\
\hline
\text{Ammonia} & 9.25 \\
\text{Acetic acid} & 4.75 \\
\text{Ammonium chloride} &...
- 19
0
votes
1
answer
119
views
How exactly does a buffer made up of a weak acid and its conjugate base work upon addition of hydroxide ions? [duplicate]
Consider a buffer solution containing a weak acid and its conjugate base ($c_\mathrm{total} = \pu{0.1 mol L-1}$):
$$\ce{HA <=> H^+ + A^-}$$
When $\ce{H+}$ ions are added (from a strong acid), ...
- 551
-1
votes
1
answer
513
views
How to calculate how much weak acid would dissociate in water?
So I know water has a pH of 7. If we add 0.1 mol of a strong acid inside 1 liter of water, it WILL fully dissociate into 0.1 mol of A- and 0.1 mol of H+ thus making the pH of the solution 1.
But what ...
5
votes
0
answers
187
views
What pH value of 1M sodium crotonate solution supposed to be?
I need to prepare the $\pu{1M}$ solution of sodium crotonate, however, I don't have one in lab, but I do have sodium hydroxide and crotonic acid.
Thus, I calculated that for $\pu{100 mL}$ of $\pu{1M}$ ...
- 151
2
votes
2
answers
7k
views
Preparation of ammonium acetate buffer for HPLC
Buffer solutions are typically prepared by dissolving an appropriate salt in water and then adjusting the $\mathrm{pH}$ to the desired position by addition of the conjugate acid or base.
For example, ...
- 29
0
votes
0
answers
52
views
Is contribution to pH by an acid or base always greater than salt hydrolysis?
I was solving a titration question and the reaction I came to was this:
$$
\begin{array}{lcccc}
\ce{&BOH(aq) &+ &HCl(aq) -> &BC(aq) &+ &H2O(l) }\\
\text{Initial} & x &...
