New answers tagged

1

Believe it or not ! If a saturated NaCl solution is added to an equal volume of $\ce{HCl}$ $0.01$ M, the pH goes from $2$ to $1.4$ ! I know : it seems contradictory, and even incredible. Diluting an acidic solution with a non-acidic solution produces a more concentrated acidic solution. Personally when I read about this experiment a long time ago, I would ...


1

If you have equal volumes of Solution A containing 0.01 M $\ce{HCl}$ and 0.1 M $\ce{BaCl2}$ and solution B containing 0.1 M apiece $\ce{Na2SO4}$ and $\ce{NaHSO4}$, each one has a pH of 2 or nearly so (based on the Henderson-Hasselbach Equation for B). If you now mix them the barium ions from A selectively precipitate the sulfate ions rather than the ...


4

There is an easy way to do what OP wants, assuming OP wants to prepare solutions at $\pu{25 ^\circ C}$. So, OP can prepare saturated solution of $\ce{Mg(OH)2}$ solution: $$\ce{Mg(OH)2_{(s)} <=> Mg^2+_{(aq)} + 2OH-_{(aq)}}$$ Since $K_\mathrm{sp}$ of $\ce{Mg(OH)2}$ is $\pu{5.61 \times 10^{-12} M3}$, you can find the solubility of $\ce{Mg(OH)2}$ at $\pu{...


1

Now much base of any kind you need to get $\mathrm{pH} = 7$? None, the water itself is enough. For $\mathrm{pH}$ levels near 7 (8 is pretty much eligible) you want to account for the $\ce{OH-}$ ions that come from water. $[\ce{OH-}] = 1 \times 10^{-6} (\mathrm{pH} = 8)$ $[\ce{OH-}][\ce{H+}] = 1 \times 10^{-14} $ (water at room temp) $[\ce{OH-}] = [\ce{H+}] + ...


4

If $c$ is the concentration of $\ce{Mg^2+}$ in the $\ce{Mg(OH)2}$ solution, the concentration $[\ce{OH-}] = 2c.$ At $\mathrm{pH}~9,$ $[\ce{OH-}] = \pu{1E-5 M},$ then $c = \pu{5E-6 M}.$ So you have to dissolve $\pu{1.25 μmol}$ $\ce{Mg(OH)2}$ in $\pu{250 mL}$ water. This is $\pu{71.3 μg}$ of $\ce{Mg(OH)2}.$ This is difficult to do in practice, as such a small ...


3

$\ce{Mg(OH)2}$ is a strong base since it is ionic in nature; it usually dissociates completely and so it's degree of dissociation is one. For weaker salts, the concentration values you assigned for the ions, $x$ and $2x$ respectively, do depend on the molarity of the $\ce{Mg(OH)2}.$ You have to define $x$ in the terms of its degree of dissociation $(\alpha),$...


3

By using a indicator, you cannot achieve $\mathrm{pH} \ \pm 0.1$ accuracy. To get that accuracy, as Karl suggested, you must buy a precision $\mathrm{pH}$ meter (you still has to calibrate it before use). The best indicator to get fairly accurate reading within the $\mathrm{pH}$ you are interested is bromothymol blue indicator. The following is a display of ...


4

Take four test tubes with sample, add bromothymol blue in two of them, methyl red in the rest. Add a few drops of HCl to one of the BTB tubes, and a few drops of NaOH in one of those with methyl red. One of the tubes will have a noticable change of colour.


3

You cannot "predict" any error by thinking that if we calibrate with two buffers the error will be x, and if we calibrate with three buffers the error will be y. Ideally, in each case, the pH of the sample should be identical. Good quality pH meters tell you how close they are to the theoretical slope in terms of percentage. It should be close to ...


3

The equation $$\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = \mathrm{p}K_\mathrm{w}\tag{1}$$ has two degrees of freedom, so two values are independent on each other and the third one depends on the other two. Reaction equilibrium constant $K_\mathrm{a}$ for $$\ce{HA + H2O <=> H3O+ + A-}\tag{R1}$$ is chemically independent on $K_\mathrm{w}$ for ...


1

The first acid dissociation constant for genistein ($\mathrm{p}K_\mathrm{a1}$) is $6.51 \pm 0.20$ (Ref.1), mainly because of hydrogen bonding between $\ce{C_{(4)}}$ carbonyl oxygen and $\ce{C_{(5)}-OH}$, similar to that in acetylacetone (see the diagram): Therefore, in physiological $\mathrm{pH}$ ($\mathrm{pH} \approx 7.3$) it should be partially ionized. ...


1

Medical saline has a $\mathrm{pH}$ of 5.5 do to the dissolved $\ce{CO2}$ within the solution as well as factoring in the degradation of the PVC packaging. I reference the attached article from the International Journal of Medical Sciences from 2013: Benjamin AJ Reddi, “Why Is Saline So Acidic (and Does It Really Matter?),” Int. J. Med. Sci. 2013, 10(6), 747-...


-1

Citric acid will not help flush Ammonium Nitrate from soil, as the other poster says, water will do that fine. For a fertiliser spill on soil, after removing whatever fertiliser you can, add sugar (sucrose) first then water. Microbial action will clean it up.


1

Efficiency of a buffer is not an absolute parameter, but it is conditional and case dependent. If buffer differencial capacity ( $\frac{\mathrm{d}n}{\mathrm{dpH}}\cdot \frac{1}{V}$, where $n$ is molar amount of an added base/acid and $V$ is buffer volume) or integral capacity ($\frac{\Delta n}{\Delta \mathrm{pH}}\cdot \frac{1}{V}$ per $\Delta pH = \pm 1$) ...


-1

Can't quite follow your equations but your friend is right. It takes twice as much $\ce{NaOH}$ as $\ce{H2SO4}$. In other words for every $\pu{1L}$ of $\pu{5M}$ $\ce{H2SO4}$, you need $\pu{2L}$ of $\pu{5M}$ $\ce{NaOH}$. The reactions are: $$\ce{2H+ + 2OH- -> 2H2O}$$ $$\ce{2Na+ + SO4^2- -> Na2SO4}$$ OVERALL: $$\ce{2NaOH + H2SO4 -> 2H2O + Na2SO4}$$


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