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Donate a proton (acting as an acid)
A covalent bond with hydrogen breaks, hydrogen leaves without electrons (as a hydrogen ion, sometimes called proton as a shorthand), and the two electrons from the bond remain with the remainder of the substance (typically in form of a lone pair).
Accept a proton (acting as a base)
A substance makes a new covalent bond ...
3
Using concentrations and not activities, user GRSousaJr presents the exact solution to the problem in equations 1-8. However the simplifications don't really seem appropriate. This is a chemistry problem, not a problem in numerical analysis to solve. The gist is that we can make some simplifications based on the number of significant figures and a knowledge ...
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How would be the equation and the ICE table, and what is the pH of the mixture of these two solutions?
There are three equations:
$$\ce{HCOOH <=> H+ + HCOO-}$$
$$\ce{CH3COOH <=> H+ + CH3COO-}$$
$$\ce{H2O <=> H+ + OH-}$$
Hydroxide is a (very) minor species, so you can neglect it. As I explain below, at the pH of the mixture, the acetate ...
1
Active carbon is neither an acid, nor a base. It cannot. It does not react directly in water and in acidic or basic solutions. But it may have a catalytic action on the solutes. It may accelerate the rate of chemical reactions. That is probably what you have observed when mixing sodium bicarbonate in water. Without knowing it, you must have increased the ...
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The question asks:
I need to create a $100\textrm{mL}$ buffer of $\textrm{pH = 9.20}$ with ammonia and ammonium chloride such that $\textrm{pH = }9.20\pm0.50$ with $20\textrm{mL}$ of $0.2\textrm{M } \ce{NaOH}/\ce{HCl}$. I am provided with $0.1\textrm{M}$ ammonia and ammonium chloride.
User Mathew Mahindaratne has provided an answer, but did not answer ...
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Your initial calculations using Henderson-Hasselbalch equation is correct:
$$
\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} \\
\mathrm{9.20} = 9.25 + \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} \\
\log \frac{[\ce{NH3}]}{[\ce{NH4+}]} = -0.05\\
\frac{[\ce{NH3}]}{[\ce{NH4+}]} = 0.89
$$
Yet, since you have only $\pu{0.10 M}$ ammonia ...
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