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Ksivusya [100]
3 years ago
9

Calculating the volume of 0.05mol/dm3 KOH is required to neutralise 25.0cm3 of 0.0150mol/dm3 HNO3

Chemistry
1 answer:
kvasek [131]3 years ago
8 0

Answer:

You first need to construct a balanced chemical equation to describe the reaction:

KOH + HNO3 ---------> KNO3 + H2O

Work out the no. moles of HNO3 being neutralized:

Moles = Volume x Concentration = (25/1000) x 0.0150 = 0.000375 moles

From the balanced equation the molar ratio of KOH to HNO3 is 1:1 so you also need 0.000375 moles of KOH to neutralise the nitric acid

Now you can work out the volume of KOH required:

Volume = Moles/Concentration = (0.000375)/0.05 = 0.0075 dm^3 = 7.5 cm^3

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A: (a) 0 (b) 20

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Answer:

a. Al(OH)_3(s)+3H^\rightarrow Al^{3+}+3H_2O(l)\\

b. Pb^{2+}(aq)+2I^-(aq)\rightarrow PbI_2(s)

Explanation:

Hello,

a. In this case, the overall reaction is:

Al(OH)_3(s)+3HBr(aq)\rightarrow AlBr_3(aq)+3H_2O

Nevertheless, the ionic version is:

Al(OH)_3(s)+3H^++3Br^-(aq)\rightarrow Al^{3+}+3Br^-(aq)+3H_2O(l)\\

Since the base is insoluble, thereby, the balanced net ionic equation turns out:

Al(OH)_3(s)+3H^\rightarrow Al^{3+}+3H_2O(l)\\

Since bromide ions become spectator ions.

b) In this case, the overall reaction is:

Pb(NO_3)_2(aq)+2LiI(aq)\rightarrow PbI_2(s)+2LiNO_3(aq)

Nevertheless, the ionic version is:

Pb^{2+}(aq)+2(NO_3)^-(aq)+2Li^+(aq)+2I^-(aq)\rightarrow PbI_2(s)+2Li^+(aq)+2(NO_3)^-(aq)

Since lead (II) iodide is insoluble whereas lithium nitrate does not, thereby, the net ionic equation turns out:

Pb^{2+}(aq)+2I^-(aq)\rightarrow PbI_2(s)

Since lithium and nitrate ions become spectator ions.

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