Answer to Question #86053 in General Chemistry for thomas

Question #86053
Oxalic acid dihydrate is a solid, diprotic acid that can be used in the laboratory as a primary standard. Its formula is H2C2O4•2H2O.

A student dissolves 0.465 grams of H2C2O4•2H2O in water and titrates the resulting solution with a solution of potassium hydroxide of unknown concentration. If 21.0 mL of the potassium hydroxide solution are required to neutralize the acid, what is the molarity of the potassium hydroxide solution?

___________ M
1
Expert's answer
2019-03-21T06:24:53-0400
H2C2O42H2O+2KOH=K2C2O4+4H2OH_2C_2O_4\cdot 2H_2O + 2KOH = K_2C_2O_4 + 4H_2O

M(H2C2O4H2O)=126.08gmolM(H_2C_2O_4\cdot H_2O) = 126.08 \frac{g}{mol}


n=mMn=\frac{m}{M}


n(C2H2O42H2O)=0.465126.08=0.00369moln(C_2H_2O_4\cdot 2H_2O) = \frac{0.465}{126.08} = 0.00369 mol


According to equation mole ratio n(C2H2O42H2O):n(KOH)=1:2n(C_2H_2O_4\cdot 2H_2O):n(KOH) = 1:2 , then


n(KOH)=2×0.00369mol=0.00738moln(KOH) = 2\times 0.00369 mol = 0.00738 mol


c=nVc = \frac{n}{V}


c=0.00738mol0.0210L=0.351molL=0.351Mc = \frac{0.00738 mol}{0.0210L} = 0.351 \frac{mol}{L} = 0.351 M


Answer: 0.351 M


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