5. Determine the pH for the following solutions a. IOH1-1.0 x 107M PHNeira [H3O+

Question

5. Determine the pH for the following solutions a. IOH1-1.0 x 107M PHNeira [H3O+)-4.2 x 10" M Pn = 2.38 AciSic [H30 0.0001 M PH= c. Calculate the volume in mL, of 0.215 M KOH solution nceded to neutralize each of the following: a. 2.50 mL of a 0.825 M H2SO4 solution b. 18.5 mL of a 0.560 MHNO, solution c. 5.00 mL of a 3.18 M HCI solution 6) Identify the conjugate acid-base pairs in each of the following equations: a. HNOdaq) + HS(aq) NO"(aq) + HgS(g) HNO /NO-on b. HCI (aq)+ OH (aa) CI(aq) +#200) /con 7. h5/HS- Ha0/0H

Explanation / Answer

6)

a)

we have the Balanced chemical equation as:

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

Here:

M(H2SO4)=0.825 M

M(KOH)=0.215 M

V(H2SO4)=2.5 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*0.825 M *2.5 mL = 1*0.215M *V(KOH)

V(KOH) = 19.2 mL

Answer: 19.2 mL

b)

we have the Balanced chemical equation as:

HNO3 + KOH ---> KNO3 + H2O

Here:

M(HNO3)=0.56 M

M(KOH)=0.215 M

V(HNO3)=18.5 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*0.56 M *18.5 mL = 1*0.215M *V(KOH)

V(KOH) = 48.2 mL

Answer: 48.2 mL

c)

we have the Balanced chemical equation as:

HCl + KOH ---> KCl + H2O

Here:

M(HCl)=3.18 M

M(KOH)=0.215 M

V(HCl)=5.0 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*3.18 M *5.0 mL = 1*0.215M *V(KOH)

V(KOH) = 74.0 mL

Answer: 74.0 mL

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