I\'ve been working thsis problem and this part is frustratingme. Can anyone help
ID: 680726 • Letter: I
Question
I've been working thsis problem and this part is frustratingme. Can anyone help? I'd appreciate it. Exactly 100 mL of 0.18 Mnitrous acid (HNO2) are titrated with a 0.18 M NaOH solution. Calculate the pH forthe following. b.)the point at which 80 mL of the base has been added I've been working thsis problem and this part is frustratingme. Can anyone help? I'd appreciate it. Exactly 100 mL of 0.18 Mnitrous acid (HNO2) are titrated with a 0.18 M NaOH solution. Calculate the pH forthe following. b.)the point at which 80 mL of the base has been addedExplanation / Answer
The number of moles of HNO2 present = M*V(in L) = 0.18 mole/L * 0.1 L = 0.018 mole Threfore the number of NaOH titrated to theneutralization = 0.18 mole/L*0.080 L = 0.0144 mole HNO2(aQ) + NaOH(aq) <-------> NaNO2(aQ) + H2O(l) 0.018 mole 0.0144mole 0.0144 mole Therefore the number ofmoles of HNO2 remained = 0.018mole - 0.0144 mole =0.0036 mole Theconcentration of the HNO2 present at equillibrium = 0.0036mole / 0.180 L = 0.02 M according to Henderson's equation pH = pKa + log[base]/[acid] = 3.36 + log(0.08/0.02) = 3.96Related Questions
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