3
Strontium hydroxide is a strong base, so you can calculate $[\ce{OH-}]$ as $\pu{0.02 M}$, then use
$$K_\mathrm{w} = \ce{[OH-][H3O+]}$$
$$1\cdot 10^{-14} = 0.02\cdot [\ce{H3O+}] \quad\to\quad [\ce{H3O+}] = \pu{5e-13 M} \quad\to\quad \mathrm{pH} = 12.3$$
3
The solubility of calcium salts is highly dependent on pH. For example, let’s look at tricalcium phosphate, which presents the solubility equilibrium:
$$\ce{Ca3(PO4)2(s) + H2O <=> 3Ca^2+(aq) + 2PO4^3-(aq)}$$
The solubility product constant for this equilibrium ($K_\mathrm{sp}$) is vastly varied from source to source. Some listed it as $\pu{2.0 × 10^–29 ...
3
Although pH scale is rather strictly used for aqueous solutions, the concept of pH in organic or non-aqueous solutions is not trivial. It is called apparent pH or operational pH when organic solvents are present. There is a practical problem here as well, do you think your buffers will dissolve in ethanol? Once you figure this out, you can read a section in ...
2
In a chemical sense, "anode" of an electrolytic cell is the most powerful oxidizing reagent known, so much so that it oxidize F(-) to elemental fluorine. Don't forget that the electron balance in an electrolytic cell is always maintained so the gram-equivalents of x reduced at the cathode must equal gram equivalents of y oxidized at the anode. As a result, ...
2
To solve this I used the Henderson-Hasselbalch equation:
$$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{[\ce{A^-}]}{[\ce{HA}]}\right)$$
Because $\ce{pH} = -\log(\mathrm{p}K_\mathrm{a})$ we end up with:
$\ce{pH} = -\log(\mathrm{p}K_\mathrm{a})+ \log \left(\frac{1}{3}\right)=-\log \left(6.7 \times 10^{-5}\right) + \log \left(\frac{1}{3}\right)= 3....
2
If we consider $\ce{HA}$ as a weak acid, then at the half equivalence point, $$\mathrm{p}H = \mathrm{p}K_\mathrm{a}$$ As $$\mathrm{p}H = \mathrm{p}K_\mathrm{a} + \mathrm{p[\ce{HA}]} -\mathrm{p[\ce{A-}]}$$ and for the half equivalence point,
$$\mathrm{p[\ce{HA}]} =\mathrm{p[\ce{A-}]}$$
So the higher the $\mathrm{p}K_\mathrm{a}$ is, the higher is $\mathrm{p}H$ ...
2
Pure water (rain as well as distilled water) in equilibrium with the atmosphere ($p_{\ce{O2}}=10^{-3.5}\ \mathrm{atm}$) can be calculated to contain about
$$\begin{align}
\mathrm{pH}=-\log[\ce{H+}]&=5.65\\
-\log[\ce{HCO3-}]&=5.65\\
-\log[\ce{CO3^2-}]&=10.3\\
-\log[\ce{H2CO3^*}]&=5.0\\
-\log[\ce{CO2}]&=5.0\\
-\log[\ce{H2CO3}]&=7.8\\
\...
1
Based on my comment above, this is what I think the table should look like.
\begin{array}{c | c c c c c c}
\text{RXN} &\ce{Zn^2+}\text{(aq)} & \ce{CH3COO-}\text{(aq)} & \ce{Na+OH-}\text{(aq)} & \ce{Zn(OH)2}\mathrm{(s)} \\\hline
\text{I} &\pu{0.05 M} & \pu{0.10 M} &\pu{ 0.025 M} &\pu{0 moles} \\
\text{...
1
At low pHs there is little "free" $\ce{CO3^{2-}}$ to precipitate the $\ce{Mg^{2+}}$ and $\ce{Ca^{2+}}$ cations. Most of the carbonate species are dissolved $\ce{CO2}$, $\ce{H2CO3}$ and $\ce{HCO3-}$. Thus the precipitate forms slowly and you get relatively large crystals.
At high pHs there is a lot "free" $\ce{CO3^{2-}}$ to precipitate the $\ce{Mg^{2+}}$ ...
1
Any good general chemistry textbook has a chapter on electrolysis. If you wish to add more scholarly work check the Journal of Chemical Education. https://pubs.acs.org/action/doSearch?text1=Electrolysis&quickLinkYear=&quickLinkVolume=&field1=Title&type=within&publication=346464552
In the first part you electrolyze a solution of salt. ...
1
Edit: At the equivalence point, the solution contains the dissolved salt $\ce{CH3NH3Cl}$, dissociated to $\ce{CH3NH3+ + Cl-}$, as @MaxW noted.
You can calculate $pH$ of the conjugate acid $\ce{CH3NH3+}$ at its dissociation equilibrium.
$$\begin{align}
K_w &= K_a . K_b = 10^{-14}\\
pK_w &= pK_a + pK_b = 14\\
\ce{ CH3NH3+ &<=> H+ + CH3NH2 }...
1
Can I multiply Ka1 and Ka1 to eliminate [$\ce{C4H5O6−}$], and then get the concentration of C4H4O62− necessary by plugging in 0.1 M for [C4H6O6] and the target pH in the appropriate form in [H+]
No, because $\ce{C4H5O6−}$ is a one of the major species. In fact, if you add the tartaric acid and its double salt at equimolar ratios, $\ce{C4H5O6−}$ will be the ...
1
Generally, the maximum buffer capacity is at $\ce{pK_a}$ . The tartaric acid is somewhat special for 2 reasons:
It is a diprotic acid with both $\ce{pK_a}$ very close, with the $\ce{pK_{a1}}$ rather low, being affected by the reason 2. :
$$\ce{pK_{a1}}=2.89,\ce{pK_{a2}}= 4.40 (L+)$$
The solution buffer capacity (not limited to presence of specific buffer ...
1
You are on the right track according to the directions given to you ( = the problem statement), but the directions themselves are wrong.
Say you had an equilibrium with $\ce{{[H^+][A^-]\over[HA]}=K}$. You dilute it five-fold and tell your particles to wait a bit with dissociation until you are done with the math. That makes all concentrations 5 times ...
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