Question

Question 8 (1 point) If there is 1.0 mole of HCN and 1.0 mole of NaCN in a 100.0 ml aqueous solution and then 0.1 mole of HCl is added what happens? Please make sure to mark all of the correct responses. The number of moles of HCN will increase. The number of moles of HCN will decrease. The number of moles of CN will increase. The number of moles of CN will decrease. The pH will increase. The pH will decrease.

Answer 1

100ml 100 Buffer: (HEN I NOCN 1.0 mole 1 + 1.0 mole - Acidic Buffer 1oome weak acid conjugate Base Now, concentration of HCN (in terms of molarity) - 1 x1000 = 10M concentration of NaCN = 1 x1000 = IOM let the dissociation constant of weak acid HCN = Ka Befcere addition of HCL; using Henderson - Hasselbalch Equection pH = pka where, [A] = concentration of conjugate base. [HA] = Concentration of weak acid [HCN] = 10M and [NaCN] = TOM 100- LHAT in pH = pka & lof (10) - pka & logo1 =pka [:loha ² = 0] ..(phpka (prior to addition) Now, when o.1 mole HCL is added to the buffer solution (100me) :. [HCL] = 0.1x1068 = 1M reaction: He & NOCN - HAN & Nace 100 Том TOM IM OM limiting reagent and IM Hal will be clearly, Hee is the completely utilised.

Ha + NaCN - HEN & Nace 10м TOM 1M OM +1M +1M -1M - 1M 9 TIM IM 11M 1M ам OM : moles = molavity x volume (L) HON moles of = 11x 100 = 1.1 moles 1000 and moles of NaCN = 9 x 100 = 0.9 mole 1000 i = The number of moles of HCN will increase. =) The number of moles of CN- will decrease. Now, using Henderson Hasselbalch Equation pH = pka + leeg LA = pka + log, A =pka + rog, 09 - loso" apka + 0.954 - 1.041 =pka 0.087 pH = pka_ 0.087 (after addition) The pH will decrease.
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