A buffer was made taht consisted of 500 mL of 0.90 MNH 3 and 500 mL of 0.90 M NH
ID: 76376 • Letter: A
Question
A buffer was made taht consisted of 500 mL of 0.90 MNH3 and 500 mL of 0.90 M NH4+. TheKa of NH4+ is 5.6 x 10-10. Calculate the reasulting pH if 0.02 mol of HCl is added to theoriginal buffer and the resulting pH if 0.02 of NaOH is added tothe original buffer. A buffer was made taht consisted of 500 mL of 0.90 MNH3 and 500 mL of 0.90 M NH4+. TheKa of NH4+ is 5.6 x 10-10. Calculate the reasulting pH if 0.02 mol of HCl is added to theoriginal buffer and the resulting pH if 0.02 of NaOH is added tothe original buffer.Explanation / Answer
The buffer solution has 0.90 MNH3 , V= 500mL 0.90 M NH4+ , V= 500mL And Ka of NH4+ =5.6*10-10 Whenever the buffer has taken , we can add to that some molsof either acid or base there is a small change in the pH of buffer observed. In this case we can excepress the pH by theHenderson-Hasselbalch equation . pH = pKa + log ([congugate base]/[acid]) So ,Now we can add 0.02 mol of HCl or 0.02mol of NaOH to the buffer solution then find the Resukting pHof solution . pH = pKa + log ([congugate base]/[acid]) = -log5.6*10-10 + log (0.90/0.90) =-log5.6*10-10 +0 pH = 9.25 Whenever the buffer has taken , we can add to that some molsof either acid or base there is a small change in the pH of buffer observed. In this case we can excepress the pH by theHenderson-Hasselbalch equation . pH = pKa + log ([congugate base]/[acid]) So ,Now we can add 0.02 mol of HCl or 0.02mol of NaOH to the buffer solution then find the Resukting pHof solution . pH = pKa + log ([congugate base]/[acid]) = -log5.6*10-10 + log (0.90/0.90) =-log5.6*10-10 +0 pH = 9.25Related Questions
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