Baking soda is pure sodium bicarbonate, NaHCO3, which dissolves and dissociates

Question

Baking soda is pure sodium bicarbonate, NaHCO3, which dissolves and dissociates in water into thesodium ion and the bicarbonate ion. As most students know from demonstrations in grade school or athome, it can be mixed with vinegar for a rapid and entertaining chemical reaction that generates carbondioxide gas bubbling out of the reaction mixture.

The overall reaction can be written as:CH3COOH (aq) + Na+(aq) + HCO3–(aq) Na+(aq) + CH3COO–(aq) + H2O(l) + CO2(g) in which the sodium ion is a spectator ion.

Baking soda or baking powder can be used in recipes as a leavening agent—that is, to produce carbondioxide gas to make dough or batter rise. Recipes using baking soda must include a separate acidicingredient such as buttermilk or vinegar. Recipes that do not include the separate addition of an acidicingredient usually require baking powder instead. Baking powder has a solid organic acid mixed in withthe baking soda, so that when the mixture is dissolved, the acid and sodium bicarbonate react with eachother. While commercial baking powder involves slightly more complicated chemistry, a good substitutefor baking powder can be made by adding baking soda to an acid such as potassium hydrogen tartrate(potassium bitartrate or cream of tartar). The formula for potassium hydrogen tartrate is KH C4H4O 6 . Your task is to devise a suitable substitute for baking powder by calculating the mass of cream of tartar in grams required to react with 10.00 g of baking soda. There should be enough cream of tartar added so that all of the baking soda will react to make carbon dioxide gas, but no more cream of tartar than that.

Explanation / Answer

The reaction between baking soda and sodium potasium tartarare is represented by

NaHCO3(aq) + KHC4H4O6(aq) KNaC4H4O6(aq) + H2O + CO2(g)   (1)

mass of baking soda= 10gms

molecular weight of baking soad (NaHCO3) can be caculated by knowing the atomic weights of elements involved

atomic weights are Na= 23, H=1,C=12 and O=16

NaHCO3 contains 1Na, 1H, 1 C and 3 Oxygen atoms and hence its atomic weight= 23+1+12+3*16= 23+1+12+48=84

moles of NaHCO3= mass/molecular weight = 10/84 moles =0.119 gmoles

From the reaction (1) moles of sodium potasium tartarate required also 0.119 moles

Molecular weight of C4H5KO6 is =4*12+ 5*1+39+6*16=188

amount of tartar =moles*molecular weight =0.119*188=22.372 gms

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