Question

In: Chemistry

1. Determine the molarity of a KOH solution when each of the following amounts of acid...

1. Determine the molarity of a KOH solution when each of the following amounts of acid neutralizes 25.0mL of the KOH solution.

5.00mL of 0.500 M H2SO4

20.00 mL of 0.250 M HNO3

13.07mL of 0.100 M H3PO4

10.00mL of 1.00 mL of 1.00 M HCl.

2 How many mEq of HCO3- are present in a solution that also contains 75 mEq of Na+, 83mEq K+, 35 mEq HCO3-, 153 mEq Cl-.

3 .What is the PH of a buffer that is 0.230 M in a weak acid and 0.500M in the acid’s conjugate base? The pka for the acid is 6.72.

4. Calculate the pKa value for each of the following acids

Nitrous acid Ka = 4.5 *10-4

Carbonic acid, Ka = 4.3 * 107

Dihydrogen phosphate ion, Ka =6.2 * 10-8

Sulfurous acid, Ka   = 1.5 *10-2

5. The HCL in a 1.10 M HCI solution is 100% dissociated. What are the molar concentrations of HCI, H3O+ and Cl- in the solution.

6.What is the PH of a buffer that is 0.150 M in a weak acid and 0.150M in the acid’s conjugates base? The acids ionization constant is 6.8 * 10-6

Solutions

Expert Solution

1)

a)

Balanced chemical equation is:

H2SO4 + 2 KOH ---> K2SO4 + 2 H2O

Here:

M(H2SO4)=0.5 M

V(H2SO4)=5.0 mL

V(KOH)=25.0 mL

According to balanced reaction:

2*number of mol of H2SO4 =1*number of mol of KOH

2*M(H2SO4)*V(H2SO4) =1*M(KOH)*V(KOH)

2*0.5*5.0 = 1*M(KOH)*25.0

M(KOH) = 0.2 M

Answer: 0.200 M

b)

Balanced chemical equation is:

HNO3 + KOH ---> KNO3 + H2O

Here:

M(HNO3)=0.25 M

V(HNO3)=20.0 mL

V(KOH)=25.0 mL

According to balanced reaction:

1*number of mol of HNO3 =1*number of mol of KOH

1*M(HNO3)*V(HNO3) =1*M(KOH)*V(KOH)

1*0.25*20.0 = 1*M(KOH)*25.0

M(KOH) = 0.2 M

Answer: 0.200 M

c)

Balanced chemical equation is:

H3PO4 + 3 KOH ---> K3PO4 + 3 H2O

Here:

M(H3PO4)=0.1 M

V(H3PO4)=13.07 mL

V(KOH)=25.0 mL

According to balanced reaction:

3*number of mol of H3PO4 =1*number of mol of KOH

3*M(H3PO4)*V(H3PO4) =1*M(KOH)*V(KOH)

3*0.1*13.07 = 1*M(KOH)*25.0

M(KOH) = 0.1568 M

Answer: 0.157 M

d)

Balanced chemical equation is:

HCl + KOH ---> KCl + H2O

Here:

M(HCl)=1.0 M

V(HCl)=1.0 mL

V(KOH)=25.0 mL

According to balanced reaction:

1*number of mol of HCl =1*number of mol of KOH

1*M(HCl)*V(HCl) =1*M(KOH)*V(KOH)

1*1.0*1.0 = 1*M(KOH)*25.0

M(KOH) = 0.04 M

Answer: 0.0400 M

Only 4 parts at a time


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