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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 ...
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}... 2 You run into all sorts of problems when the pH is outside of the range of 1 to 13. For example, in Environ. Sci. Technol. (2000) 342, p. 254-258, they say: [...] the former National Bureau of Standards (NBS) established a set of conventions that limits measurements to 1 < pH < 13 and to ionic strength, I < 0.1. The main limitations are the ... 2 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 ... 1 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 ...