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I'm looking for some way to calculate how much base I will need to add to an unbuffered solvent consisting of a mixture of ethanol and water in order to basify it to a specific reading (10) on an a pH probe calibrated with an aqueous standard. I think you have summarized the main issue yourself. The reading of a pH probe in a basic alcoholic solution is ...


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Your solution is correct up to the point you assumed that you can double the concentration of hydrochloric acid. Unfortunately, this is wrong and not at all how stoichiometry works. Let's focus on what's important. In the nutshell, we are dealing with a typical neutralization reaction: $$\ce{H3O+ + OH- <=> 2 H2O}$$ and note that $\ce{BaCl2}$ as a ...


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You can look at the question this way: when we have a strong acid/strong base titration, at the equivalence point we would end up with a neutral solution (the $\mathrm{pH}$ would be ~$7$). If we had a weak acid/strong base titration, at the equivalence point we would end up with $\mathrm{pH}$ above 7 (basic solution has $\mathrm{pH}$ between $8-10$). If ...


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Pure water dissociate according to the equation: $$\ce{2H2O(l) <=> H3O+(aq) + OH- (aq)}$$ Assume that $K_\mathrm{w} = 1.00 \times 10^{-14}$ at room temperature. Thus, $\ce{[H3O+]}$ of solution is $\pu{1.00E-7 M }$. Now, if you add a trace amount of strong acid, this equilibrium would be disturbed and backward reaction occur to reduced some of added ...


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Suppose you have a $\pu{1.0 L}$ of acetate buffer solution made by $\pu{500.0 mL}$ of $\pu{1.0 M}$ $\ce{CH3COONa}$ solution adding to $\pu{500.0 mL}$ of $\pu{1.0 M}$ $\ce{CH3COOH}$ solution. Thus, $[\ce{CH3COOH}] = [\ce{CH3COONa}] = \pu{0.5 M}$. Therefore, according to Henderson–hasselbalch equation, $\mathrm{pH} = \mathrm{p}K_\mathrm{a} = 4.75$ ($\mathrm{p}...


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The question This is the question, corrected according to Poutnik's comment, and adding (i) and (ii) to distinguish the two scenarios: $\ce{NH3}$ solution of $\pu{0.1 mol dm-3}$ is being added to a $\pu{25.0 cm3}$ of $\pu{0.1 mol dm-3}$ $\ce{HCl}$ solution. Calculate the pH of the solution when volume of added $\ce{NH3}$ solution is (i) $\pu{25.0 ...


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Yes, concentrated strong acid can destroy litmus paper. Paper is mostly cellulose, which is a carbohydrate polymer. Concentrated sulfuric acid will pull the hydrogen and oxygen out of it as water, leaving carbon behind. The indicator dye that's supposed to change color probably won't fare too well, either. You can dilute the unknown by adding a drop or two ...


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Liquid pH testing solutions or drops are also available. Other indicators like phenolphthalein, methyl orange , methyl red can be used if you expect your test solution to fall in the pH range of the respective indicators. Also it will be wise to take a small sample of the solution to test it rather than testing the whole thing.


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Another approach, I believe, is this: Definitions $$\mathrm{pH} = \frac{1}{2}\left(\log K_\mathrm{b} - \log K_\mathrm{a} - \log K_\mathrm{w}\right) \\ 2\,\mathrm{pH}=\log(\frac{K_b}{K_a\,K_w})$$ It suggests equations' sum, like this: $$\ce{HA <=> A- + H+} \tag{$\ce{K_a=\frac{[A-][H+]}{[HA]}}$}$$ $$\ce{BOH <=> B+ + HO-} \tag{$\ce{K_b=\frac{[...


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