Where did they get Naoh from? 25.00 mL of a 0.100 M solution of formic acid (HCO

Question

Where did they get Naoh from?

25.00 mL of a 0.100 M solution of formic acid (HCOOH, K 1.7 x 104) is titrated with a 0.100 M solution of KOH, a strong base. What is the pH after 10.00 mL of KOH solution has been added? Moles HCOOH = 0.02500 L (0.100 mol/L) = 2.50 x 10-3 mol Moles NaOH = 0.01000 L (0.100 mol/L) 1.00 x 10-3 mol Volume = 25.00 mL + 10.00 mL = 35.00 mL HCOOH(aq) + NaOH(aq) Na+(aq) + HC00(aq) + H2O(1) Initial moles Moles after Concentration after lizationneutralization 2.50 x 103 HCOOH2 HCOO 1.50 x 103 1.00 x 103 0.04286 0.02857 1.00 x 103

Explanation / Answer

moles of formic acid = 25 x 0.1 / 1000 = 2.5 x 10^-3

moles of KOH = 10 x 0.1 / 1000 = 1.0 x 10^-3

HCOOH      +       KOH     ------------------> HCOOK    +   H2O

2.5 x 10^-3       1.0 x 10^-3                             0                 0

- 1.0 x 10^-3     - 1.0 x 10^-3                      1.0 x 10^-3    

1.5 x 10^-3             0                                  1.0 x 10^-3

pH = pKa + log [salt / acid]

    = 3.77 + log [1 x 10^-3 / 1.5 x 10^-3]

pH = 3.59

Note : they taken mistakenly instead of KOH , they took NaOH. i did clearly.

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