In: Chemistry
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
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
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