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# A student plans to titrate 25.00 mL of a .098 M HCOOH solution with .227 NaOH solution.

A student plans to titrate 25.00 mL of a .098 M HCOOH solution with .227 NaOH solution. Calculate the volume of NaOH (in mL) that will be needed to reach the equivalence point and calculate the anticipated pH at the equivalence point.

Neutralization reaction is HCOOH + NaOH ...........&gt; HCOONa + H2O. one mole... View the full answer

• at equivalence point, [HCOONa] = (0.025 * 0.098) * 1000 / (25 + 10.8) = 0.0684 M. PH = 0.5 * Pkw + 0.5 * Pka + 0.5 * log C = 0.5 * 14 + 0.5 * 4.74 + 0.5 * log (0.0684) = 8.21. so PH = 8.21
• tutor_007
• Apr 02, 2018 at 11:52pm
• sorry correction please: Pka of HCOOH = 3.8. so PH = 0.5 * 14 + 0.5 * 3.8 + 0.5 * log (0.0684) = 8.21. so PH = 7.74. Thank you.
• tutor_007
• Apr 02, 2018 at 11:55pm
• PH = 0.5 * 14 + 0.5 * 3.8 + 0.5 * log (0.0684) = 7.74. so PH of the solution is 7.74.
• tutor_007
• Apr 02, 2018 at 11:56pm

• Can I have exact value of pH ? The problem i have for this question is calculating the pH
• st216427
• Apr 03, 2018 at 12:06am
• ok... just a minute please.
• okiyaiomuse
• Apr 03, 2018 at 12:10am
• molarity of HCOONa =( molarity of NaOH * molarity of HCOOH) /( volume of NaOH + volume of HCOOH). which is (0.225*0.098) / (10.79+25) = 0.000616M. then le the molarity of HCOONa be C, we can calculate the equivalent pH. therefore pH = 1/2 [ pKw + pKa + log C ]. but Ka of HCOOH acid = 1.77*10^-4 therefore we can calculate the pKa, pKa = -log Ka = -log(1.77*10^-4) = 3.75 ; pKw = -log Kw = -log(1.0*10^-14) = 14 ; therefore equivalent pH = 0.5[ 14+3.75+log0.000616] = 0.5[14+3.75-3.21] = 0.5[14.54] = 7.27, therefore the exact pH is 7.27
• okiyaiomuse
• Apr 03, 2018 at 12:53am
• okiyaiomuse
• Apr 03, 2018 at 12:56am

acid = HCOOH base = NaOH Macid * Vacid = Mbase * Vbase 0.098 * 25.0 = 0.227 *  V base V base  = (0.098 * 25.0) / 0.227 V... View the full answer

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