From 20192f73c7bb0c318eddc27fe9da4199a5d056c9 Mon Sep 17 00:00:00 2001 From: =?utf8?q?Michal=20Mal=C3=BD?= Date: Fri, 16 Oct 2015 16:44:43 +0200 Subject: [PATCH] =?utf8?q?Pou=C5=BE=C3=ADt=20vhodn=C4=9Bj=C5=A1=C3=AD=20cN?= =?utf8?q?aA=20m=C3=ADsto=20cHA=20ve=20v=C3=BDpo=C4=8Dtu=20pH=20octanu=20s?= =?utf8?q?odn=C3=A9ho?= MIME-Version: 1.0 Content-Type: text/plain; charset=utf8 Content-Transfer-Encoding: 8bit --- tzach_problems_solved.tex | 12 ++++++------ 1 file changed, 6 insertions(+), 6 deletions(-) diff --git a/tzach_problems_solved.tex b/tzach_problems_solved.tex index 14f2b1c..e5e73f6 100644 --- a/tzach_problems_solved.tex +++ b/tzach_problems_solved.tex @@ -573,7 +573,7 @@ \paragraph{Zadání} Odvoďte odpovídající všeobecnou Brønstedtovu rovnici a s jejím použitím vypočítejte pH roztoku octanu sodného o koncentraci \num{1e-2}~\mpdm. \\ \begin{tabular}{l>{=}cr} - \ilm{pK_a(CH_3COOH)} && \num{4,75} + \ilm{pK_A(CH_3COOH)} && \num{4,75} \end{tabular} @@ -600,23 +600,23 @@ [OH^-] = K_B \frac{[A^-]}{[HA]} } \rmm{ - [OH^-] = K_B \frac{c_{HA} - \cfbox{darkgreen}{\ilm{[OH^-] - \textcolor{redorange}{[H_3O^+]}}}}{[OH^-] - \textcolor{redorange}{[H_3O^+]}} + [OH^-] = K_B \frac{c_{NaA} - \cfbox{darkgreen}{\ilm{[OH^-] - \textcolor{redorange}{[H_3O^+]}}}}{[OH^-] - \textcolor{redorange}{[H_3O^+]}} } \infloattext{\centering Platí stejná zjednodušovací pravidla jako všude jinde} \rmm{ - [OH^-]^2 = K_B c_{HA} + [OH^-]^2 = K_B c_{NaA} } \rmm{ - \textcolor{skyblue}{\frac{K_W}{[H_3O^+]^2}} = \textcolor{skyblue}{\frac{K_W}{K_A}}c_{HA} + \textcolor{skyblue}{\frac{K_W}{[H_3O^+]^2}} = \textcolor{skyblue}{\frac{K_W}{K_A}}c_{NaA} } \rmm{ - [H_3O^+] = \sqrt{\frac{K_W K_A}{c_{HA}}} + [H_3O^+] = \sqrt{\frac{K_W K_A}{c_{NaA}}} } } \FloatJail{calculation}{Výpočet pH octanu sodného ze zadání}{ \begin{align*} - [H_3O^+] &= \sqrt{\frac{K_W K_A}{c_{HA}}} \\ + [H_3O^+] &= \sqrt{\frac{K_W K_A}{c_{NaA}}} \\ [H_3O^+] &= \sqrt{\frac{10^{-14} \cdot 10^{\num{-4,75}}}{\num{1e-2}}} \\ pH &= \num{8,375} \end{align*} -- 2.43.5