A solution of malonic acid H2C3H2O4, was standardized by titration with .1000M NaOH solution. If 20.76mL of the NaOH solution were required to neutralize completely 13.15mL of the malonic acid solution, what is the molarity of the malonic acid solution?

0.079M

Explanation:

Step 1:

Writing a balanced equation for the reaction. This is illustrated below:

H2C3H2O4 + 2NaOH —> C3H2Na2O4 + 2H2O

From the balanced equation above, the following were obtained obtained:

Mole of the acid (nA) = 1

Mole of the base (nB) = 2

Step 2:

Data obtained from the question. This includes:

Molarity of the base (Mb) = 0.1M

Volume of the base (Vb) = 20.76mL

Molarity of the acid (Ma) =?

Volume of the acid (Va) = 13.15mL

Step 3:

Determination of the molarity of the acid.

Applying the equation:

MaVa/MbVb = nA/nB

The molarity of the acid can be obtained as follow:

MaVa/MbVb = nA/nB

Ma x 13.15 / 0.1 x 20.76 = 1/2

Cross multiply to express in linear form

Ma x 13.15 x 2 = 0.1 x 20.76

Divide both side by 13.15 x 2

Ma = (0.1 x 20.76) / 13.15 x 2

Ma = 0.079M

Therefore, the molarity of the acid is 0.079M

The molarity of the malonic acid solution is 0.0789 M

Explanation:

Step 1: data given

Molarity of NaOH = 0.1000 M

Volume of NaOH = 20.76 mL = 0.02076 L

Volume of malonic acid solution = 13.15 mL = 0.01315 L

Step 2: The balanced equation

C3H4O4 + 2NaOH → 2H2O + Na2C3H2O4

Step 3: Calculate the molarity of malonic acid

b*Ca*Va = a*Cb*Vb

⇒with b = the coefficient of NaOH = 2

⇒with Ca = the concentration of C3H4O4 = TO BE DETERMINED

⇒with Va = the volume of C3H4O4 = 0.01315 L

⇒with a= the coefficient of C3H4O4 = 1

⇒with Cb = the concentration of NaOH = 0.100 M

⇒with Vb = the volume of NaOH = 0.02076 L

2 * Ca * 0.01315 = 0.100 * 0.02076

Ca = 0.100 * 0.02076 / ( 2 * 0.01315)

Ca = 0.0789 M

The molarity of the malonic acid solution is 0.0789 M

i think 4.44 moles

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"