Translation: To titrate a 15ml sample of apple juice a student need 10.8ml of a

Question

Translation: To titrate a 15ml sample of apple juice a student need 10.8ml of a 0.0958M NaOH solution. Calculate: The molarity of an acidic solution The mass of the citric acid present in the sample The mass of citric acid / mL of juice To titrate a 15ml sample of apple juice a student need 10.8ml of a 0.0958M NaOH solution. Calculate: The molarity of an acidic solution The mass of the citric acid present in the sample The mass of citric acid / mL of juice Al titular una muestra de 15.00mL de jugo de manzana, un estudiantes ne 10.80mL de una solucion de NaoH O.0958M. Calcule: a) la molaridad de la solucion écide b) la masa del acido citrico presente en la muestra c) la masa del acido citrico/mL de jugo

Explanation / Answer

Basicity of citric acid=3

Thus,

The molarity of citric acid will be (1/3) of the molarity of H+ ions

Let molarity of H+ ions= M1

Volume of sample=15mL=V1

Molarity of NaOH=0.0958M=M2

Volume of NaOH = 10.8mL=V2

M1*V1=M2*V2

Thus,

M1=10.8*0.0958/15=0.0690M

(a) Molarity of citric acid as concluded earlier is= M1/3=0.023M

Moles of citric acid in sample=molarity*volume=0.023mol/L*0.015L=3.45*10-4moles

Mass of citric acid=moles*molecular weight

molecular weight of citric acid=192.12g/mol

(b) Mass of citric acid=3.45*10-4*192.12g=0.0663g

(c) mass per mL=0.0663/15=0.004419g/ml

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