Question

Calculate the concentrations of hydroxide and hydronium ions in a solution prepared by dissolving 0.471 g of solid NaOH in enough water to make 225 mL of solution. Kw = 1.0E-14.

Hydroxide ion concentration? M The number of significant digits is set to 3; the tolerance is +/-4% LINK TO TEXT

Hydronium ion concentration? M The number of significant digits is set to 3; the tolerance is +/-4%

Answer 1

Step1 : Calculate concentration of NaOH

Step2: Concentration of NaOH will be equal to hydroxide ion

Step3 : [H3O+][OH-] = Kw ; by using this we can find out hydronium ion concentration.

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Mass (of solute) given =0.471 g

Molar mass of NaOH = 39.997 g/mol

Volume = 225 mL = 0.225 L

Concentration of NaOH in terms of Molarity -

volume Molarity Molarity No. of moles of salute solution (2) Mass giver/malar mass volume of solution (2) 0.4719/-39.999 g/mol Molarity 0,225L Molarity 3. 6 5 23332 số ( ) =[Na0+ Naolt NattoH" So, [o4 0.052337258 (m) Now, [OH-] [Azot] = kw= 1x10-14 -3 + [H30+] = 1x1014 1.910684713 X 10 (M) 0052337258

NaOH and OH- share 1:1 ratio. Hence concentration of NaOH will be equal to OH- ions.

Answer up to 3 significant figures -

Hydroxide ion concentration = 0.0523 M

Hydronium ion concentration = 1.91 × 10^-13 M

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