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A chemistry student wants to create a buffer solution with a pH of 7.2 using acetic acid and sodium acetate. The student has calculated that they need 0.1 moles of each component to make a 250 mL buffer solution. However, the student accidentally added 0.02 moles of hydrochloric acid to the buffer solution. Calculate the new pH of the buffer solution and explain if t

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To calculate the new pH of the buffer solution, we will use the Henderson-Hasselbalch equation:pH = pKa + log [A-]/[HA] where pH is the desired pH, pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base  sodium acetate , and [HA] is the concentration of the weak acid  acetic acid .First, we need to find the pKa of acetic acid. The Ka of acetic acid is 1.8 x 10^-5 . To find the pKa, we take the negative logarithm of Ka:pKa = -log Ka  = -log 1.8 x 10^-5    4.74Now, let's find the initial concentrations of acetic acid and sodium acetate in the buffer solution. Since we have 0.1 moles of each component in a 250 mL solution, we can calculate the concentrations as follows:[HA] = moles of acetic acid / volume in liters = 0.1 moles / 0.25 L = 0.4 M[A-] = moles of sodium acetate / volume in liters = 0.1 moles / 0.25 L = 0.4 MWhen the student accidentally added 0.02 moles of hydrochloric acid  HCl  to the buffer solution, the HCl reacts with the sodium acetate  conjugate base  to form acetic acid and sodium chloride:HCl + CH3COO-  CH3COOH + Cl-Since 0.02 moles of HCl were added, 0.02 moles of sodium acetate will react, and 0.02 moles of acetic acid will be formed. We can now calculate the new concentrations of acetic acid and sodium acetate:New [HA] =  0.4 M + 0.02 moles / 0.25 L  = 0.48 MNew [A-] =  0.4 M - 0.02 moles / 0.25 L  = 0.32 MNow we can use the Henderson-Hasselbalch equation to find the new pH:pH = pKa + log [A-]/[HA]  = 4.74 + log 0.32/0.48   4.74 - 0.4 = 4.34The new pH of the buffer solution is 4.34. The addition of hydrochloric acid has shifted the equilibrium position of the reaction towards the acid  acetic acid , as the pH has decreased from 7.2 to 4.34.

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