1. An aqueous solution contains 0.395 M hydrocyanic acid. Calculate the pH of the solution after the addition of 5.99×10-2 moles of sodium hydroxide to 255 mL of this solution. (Assume that the volume does not change upon adding sodium hydroxide)
pH =
2. An aqueous solution contains 0.302 M acetic acid. Calculate the pH of the solution after the addition of 2.56×10-2 moles of potassium hydroxide to 125 mL of this solution. (Assume that the volume does not change upon adding potassium hydroxide)
pH =
1. no of moles of HCN = molarity * volume in L
= 0.395*0.255 = 0.100725moles
Pka of HCN = 9.31
------------- HCN(aq) + NaOH(aq) -------------> NaCN(aq) + H2O(l)
I------------0.100725 -----0.0599 ----------------------0
C--------- -0.00599 ------ -0.0599 -------------------- 0.0599
E------- 0.040825 ------- 0 -----------------------------0.0599
PH = PKa + log[NaCN]/[HCN]
= 9.31 + log0.0599/0.040825
= 9.31 + 0.1665
= 9.4765 >>>>answer
2.
no of moles of CH3COOH = molarity * volume in L
= 0.302*0.125 = 0.03775moles
Pka of CH3COOH = 4.74
------------- CH3COOH(aq) + KOH(aq) -------------> CH3COOKaq) + H2O(l)
I------------0.03775 ----- 0.0256 --------------------------------0
C--------- -0.0256 ------ -0.0256 --------------------------- 0.0256
E------- 0.01215------- 0 ------------------------------------0.0256
PH = PKa + log[CH3COONa]/[CH3COOH]
= 4.74 + log0.0256/0.01215
= 4.74 + 0.3236
= 5.0636 >>>>answer
1. An aqueous solution contains 0.395 M hydrocyanic acid. Calculate the pH of the solution after...
1. An aqueous solution contains 0.395 M hydrocyanic acid. Calculate the pH of the solution after the addition of 5.99×10-2 moles of sodium hydroxide to 255 mL of this solution. (Assume that the volume does not change upon adding sodium hydroxide) pH = 2. An aqueous solution contains 0.302 M acetic acid. Calculate the pH of the solution after the addition of 2.56×10-2 moles of potassium hydroxide to 125 mL of this solution. (Assume that the volume does not change...
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