So far I have all of the answers correct except for the last box which I left bl
ID: 887007 • Letter: S
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So far I have all of the answers correct except for the last box which I left blank. I don't understand how to get that last answer. I would like someone to please explain to me step by step how to get that last answer.
The reaction of the weak acid HCN with the strong base KOH is: HCN/aq) + KOH(aq) HOH(1) KCN(aq need to start with the stoichiometry. Let's do just the stoich in steps To compute the pH of the resulting solution if 57mL of 0.55M HCN is mixed with 15mL of 0.31 KOH we Number Number How many moles of acid? 1.03135 How many moles of base? 00465 What is the limiting reactant? KOH Number How many moles of the excess reactant after reaction? .0267 Number What is the concentration of the excess reactant after reaction?3708 Number What is the concentration of the pH active product after reaction? These are the initial concentration in an ICE Table! Incorrect.Explanation / Answer
HCN + KOH ---------------------> H2O + KCN
excess reactant (HCN) moles = 0.0267 up to this you are correct
concentration = moles / total volume
here total volume = 57 + 15 = 72 ml
we have to calculate concentration here for 72 ml not for 1 litre . that means 0.0267 moles present in 72 ml but not present in 1 litre.
concentration of excess reagent = 0. 0267 / 72
concentration of excess reagent = 3.708 x 10^-4 ----------------------------> answer
active product in this reaction is KCN .
KCN moles = KOH moles = 0.00465
KCN concentration = 0.00465 / total volume
= 0.00465 / 72
C = 6.46 x 10^-5
KCN is a salt of strong base and weak acid . so the pH should be more than 7
any weak acid + strong base type salt pH formula :
pH = 7 + 1/2 [pKa + log C]
pKa of HCN = 9.21
pH = 7 + 1/2 (9.21 + log (6.46 x 10^-5 ))
pH = 9.51 --------------------------------------------------------> answer
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