If 65 ml of a 0.430 M NaOH solution is diluted to 500 mL, what is the concentrat
ID: 480654 • Letter: I
Question
If 65 ml of a 0.430 M NaOH solution is diluted to 500 mL, what is the concentration of the diluted solution? 0.215 M 0.30 M 0.023 M 3.3 m 0.056 M The net ionic equation for the reaction between aqueous nitric acid and aqueous sodium hydroxide is H^+ (aq) + HNO_3 (aq) + 2OH (aq) rightarrow 2H_2O (l) + NO_3^- (aq) HNO_3 (aq) + NaOH (aq) rightarrow NaNO_3 (aq) + H_2O (l) H^+ (aq) + OH^- (aq) rightarrow H_2O(l) HNO_3 (aq) + OH^- (aq) rightarrow NO_3^- (aq) + H_2O(1) H^+ (aq) + Na^+ (aq) + OH (aq) rightarrow H_2O (1) + Na^+ (aq)Explanation / Answer
Answer No. 24
Based on the dilution formula of M1V1 = M2V2. Here in this case, M1 = 0.430 M ; V1 = 65 mL ; M2 = Unknown and V2 = 500 mL. Thus substituing the numbers in the above given equation results in
0.430 M X 65 mL = x X 500 mL (Here x = unknown molarity, which is to be calculated)
solving it, gives x X 500 mL = 27.95 ml.M; thus, x = 27.95 ml.M/500 mL = 0.055 M
Therefore, 0.055M is the concentration of the diluted solution of NaOH.
Answer No. 25
The answer is (C).
Reason : - Nitric acid is strong acid and Sodium hydroxide is strong base. So, its an acid-base reaction. So, here H+ and OH- reacts to give water
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