Let us consider the titration of 25.0 mL of 0.100 M acetic acid (a weak acid) with 0.100 M sodium hydroxide and compare the titration curve with that of the strong acid. The titration of a weak acid with a strong base (or of a weak base with a strong acid) is somewhat more complicated than that just discussed, but it follows the same general principles. You may try to follow methods described in the lecture on polyprotic acids and bases pH calculation, or you may use BATE - pH calculator. Therefore, we will use the quadratic formula to solve for x: ${K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{F}}^{\text{-}}\right]}{\left[\text{HF}\right]}=\frac{\left(1.0\times {10}^{-7}+x\right)x}{1.0\times {10}^{-6}-x}=7.2\times {10}^{-4}$, x2 + 7.201 $\times$ 10−4x − 7.2 $\times$ 10−10 = 0, $\begin{array}{ll}x\hfill & =\frac{-7.201\times {10}^{-4}\pm \sqrt{{\left(7.201\times {10}^{-4}\right)}^{\text{2}}-4\left(1\right)\left(-7.2\times {10}^{-10}\right)}}{2}\hfill \\ \hfill & =\frac{-7.201\times {10}^{-4}\pm 7.22097\times {10}^{-4}}{2}=9.98\times {10}^{-7}\hfill \end{array}$. The choice of an indicator for a given titration depends on the expected pH at the equivalence point of the titration, and the range of the color change of the indicator. Acid-base indicators are either weak organic acids or weak organic bases. When the base solution is added, it also dissociates completely, providing OH− ions. Find the pH after 12.50 mL of the NaOH solution has been added. In the reaction the acid and base react in a one to one ratio. 8.364 1 50mmol Salt = 150mL = 3 log pH solution:ż [pKCO + pKa +logC]orz[14 + 3.75 + logą Answer: 8.6364 8) What is the pH after adding 100 mL of NaOH? The change in concentrations is: Putting these values in the equilibrium expression gives: ${K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{F}}^{\text{-}}\right]}{\left[\text{HF}\right]}=\frac{\left(x\right)\left(x\right)}{{10}^{-2}-x}=7.2\times {10}^{-4}$, x2 + 7.2 $\times$ 10−4x − 7.2 $\times$ 10−6 = 0, $\begin{array}{ll}x\hfill & =\frac{-7.2\times {10}^{-4}\pm \sqrt{{\left(7.2\times {10}^{-4}\right)}^{\text{2}}-4\left(1\right)\left(-7.2\times {10}^{-6}\right)}}{2}\hfill \\ \hfill & =\frac{-7.2\times {10}^{-4}\pm 5.415\times {10}^{-3}}{2}=2.4\times {10}^{-3}\hfill \end{array}$. lectures the relevant material can be found. By the end of this module, you will be able to: As seen in the chapter on the stoichiometry of chemical reactions, titrations can be used to quantitatively analyze solutions for their acid or base concentrations. (17-11&12, 17-28) 3.14 = Molar mol of HCOOH 100 * 0.5 = 50 m mol m mole of NaOH = 10 ml * 1.0 = 10 m mol The reaction is HCOOH + NaOH + HCOONa + H20 salt pH = PKa + log where pka = 3.74 acid 10mmol salt acid 40mmol 110mol 110mol 10mmol pH = 3.74 + log = pH = 3.74 + log 1 110moll | or 3.74 +log [] pH = 3.74 + log [0.25] 3.74 + (-0.6020) or pH = 3.138 Ans:pH = 3.14 What is the initial pH before any amount of the NaOH solution has been added? View desktop site. This is past the equivalence point, where the moles of base added exceed the moles of acid present initially. The color change intervals of three indicators are shown in Figure 3. Answer: (1) 0.00 mL: 2.37; (2) 15.0 mL: 3.92; (3) 25.00 mL: 8.29; (4) 30.0 mL: 12.097. Calculate pH at the equivalence point of formic acid titration with NaOH, assuming both titrant and titrated acid concentrations are 0.1 M. pK a = 3.75. If we add base, we shift the equilibrium towards the yellow form. The titration curve shown in Figure 3 is for the titration of 25.00 mL of 0.100 M CH3CO2H with 0.100 M NaOH. The difference occurs when the second acid reaction is taking place. Table 1 shows data for the titration of a 25.0-mL sample of 0.100 M hydrochloric acid with 0.100 M sodium hydroxide. Solving for x gives 2.26 $\times$ 10−6M. The pH at the equivalence point is also higher (8.72 rather than 7.00) due to the hydrolysis of acetate, a weak base that raises the pH: After the equivalence point, the two curves are identical because the pH is dependent on the excess of hydroxide ion in both cases. A titration is carried out for 25.00 mL of 0.100 M HCl (strong acid) with 0.100 M of a strong base NaOH the titration curve is shown in Figure 1. There is initially 100 mL of 0.50 M formic acid and the concentration of {eq}NaOH {/eq} is 1.0 M.. A. The reaction can be represented as: ${K}_{\text{b}}=\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]}{{K}_{\text{a}}}=\frac{{K}_{\text{w}}}{{K}_{\text{a}}}=\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}=5.6\times {10}^{-10}$. Let us denote the concentration of each of the products of this reaction, CH3CO2H and OH−, as x. Recognizing that the initial concentration of HF, 1 $\times$ 10−7M, is very small and that Ka is not extremely small, we would expect that x cannot be neglected. Find the pH after 25.00 mL of the NaOH solution have been added. We’re going to titrate formic acid (HCO 2 H) with the strong base NaOH, and follow its titration curve. The equivalence points of both the titration of the strong acid and of the weak acid are located in the color-change interval of phenolphthalein. 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## formic acid and naoh titration

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