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Question: Determine the $\mathrm{pH}$ of the solution resulting when $\pu{100 cm^3}$ of $\pu{0.50 mol dm-3}$ $\ce{CH2ClCOOH}$ is mixed with $\pu{200 cm^3}$ of $\pu{0.10 mol dm-3}$ $\ce{NaOH}$. I'm not sure what level of chemistry is OP's in, but the given solution for the question is for the chemistry students with appreciable knowledge of stoichiometry of ...


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$\ce{CH2ClCOOH}$ is a weak acid; $\ce{NaOH}$ is a strong base. Now, tell me, what do you think happens when you mix a strong base with a weak acid? They react. $$\ce{CH2ClCOOH + NaOH -> CH2ClCOONa + H2O} $$ A neutralisation reaction in aqueous medium depends upon the number of equivalents of acid and base. $$N_1V_1=N_2V_2 $$ At this stage, an equivalence ...


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The R code equivalent of Poutnik's answer, for refrence: # Sumamente facil Ka.1 <- 7.1 * 10^-3 Ka.2 <- 6.3 * 10^-8 Ka.3 <- 4.5 * 10^-13 Kw <- 10^-14 P_ca <- 1 pH.seq <- seq(from=0,to=14,length.out = 1000) a1 <- function(H) H^3 a2 <- function(H) H^2 * Ka.1 a3 <- function(H) H * Ka.1 * Ka.2 a4 <- function(H) Ka.1 * Ka.2 * Ka.3 ...


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Much easier is to calculate the inverse function $[\ce{Na+}]=f([\ce{H+}],K_\mathrm{a1},K_\mathrm{a2},K_\mathrm{a3})$. Calculate fractions of respective phosphate forms as the function of [H+] and acidity constants. First, calculate the common denominator CD: $$a_1 = [\ce{H+}]^3$$ $$a_2 = [\ce{H+}]^2 \cdot K_\mathrm{a1} = a_1 \cdot \frac {K_\mathrm{a1}}{[\ce{...


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In order for you to carry out this reaction, you need potassium permanganate, which provides the ion $\ce{MnO4^-}$. You also need sulfuric acid, and quite a lot of this acid, because the equation requires $\ce{8 H^+}$. And well ! $8$ is much ! So the reaction will not work without acid. And you still need $5$ electrons, which are provided by $\ce{5 Fe^{2+}}$...


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If $1$ liter $\ce{HCl} \pu{1 M}$ is gradually neutralized by adding $\pu{x mol } \ce{NaOH}$, without change of volume, the pH of the obtained solution is given by : $$\mathrm{pH} = - \log(\frac{1-x}{2} + \frac{1}{2} \sqrt{(1-x)^2 + 4·10^{-14}})$$


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