Consider a buffer solution that contains 0.50 MNaH 2 PO 4 and 0.20 MNa 2 HPO 4 .

Question

Consider a buffer solution that contains 0.50 MNaH2PO4 and 0.20 MNa2HPO4.
a. Calculate the pH of this buffersolution. b. Calculate the pH after 0.120 g ofsolid NaOH is added to 150 mL of this buffer solution. c. If the acceptable buffer range ofthe buffer solution is ± 0.10 pHunits, calculate how many moles of H3O+ canbe neutralized by 250 mL of the buffer. Consider a buffer solution that contains 0.50 MNaH2PO4 and 0.20 MNa2HPO4.
a. Calculate the pH of this buffersolution. b. Calculate the pH after 0.120 g ofsolid NaOH is added to 150 mL of this buffer solution. c. If the acceptable buffer range ofthe buffer solution is ± 0.10 pHunits, calculate how many moles of H3O+ canbe neutralized by 250 mL of the buffer.

Explanation / Answer

a) with henderson-hasselbalch, and pKa=7.20 (textbook) pH=pka+ log (base/acid) pH=7.20 + log (0,2/0,5) pH=6.80 b) moles of NaOH = m/mw= 0,120g/40.00g/mol=0.00300 mol moles H2PO4-=C*V=0.0750 mol moles HPO4--=C*V= 0,0300 mol NaOH will react with acid and form base (you can do an ice table to help you) you now have 0,0720 mol H2PO4 and 0,00330 mol HPO4 you use henderson-hasselbalch just like in a) (ps you can plugin moles instead of concentration, because volumes are the same andwill cancel themselves out) pH= 7.20 + log (0,033/0,0720) pH=6.86 c)because you want to neutralize an acid, the pH will go down(see opposite in b)) so you use pH found in a) minus 0,10= 6.70 so now you have: 6.70 = 7.20 + log (((.2*.250)-x)/(0,5*.250)+x) ***the base neutralize acid, and forms the acid form inbuffer*** isolate x... and there's your answer!

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