A 100.0-mL buffer solution is 0.175 M in HClO and 0.150 M in NaClO. Part A: What is the initial pH of this solution?Part B: What is the pH after addition of 150.0 mg of HBr?Part C: What is the pH after addition of 85.0 mg of NaOH?

Initial pH of this pH  = 7.453

pH after addition of 150.0 mg of HBr  =  7.35

pH after addition of 85.0 mg of NaOH 0.154  = 7.56

Explanation:

Since Ka value isn't given  

so we use Ka value of HClO (hypo chlorous acid) = 3 x 10⁻⁸

pKa = - logKa = 7.52

Part A

Using Henderson equation

pH = pka + log

pH = 7.52 + log

pH = 7.453

Part B

pH after addition of 150 mg of HBr

moles of HBr    

Moles of NaOCl in 100 ml buffer solution = = 0.015

Moles of HClO in 100 ml buffer solution = = 0.0175

Since H⁺ concentration furnished by HBr acid make a common ion effect . So the following reaction carried out

ClO⁻ + H⁺ → HClO

So the remaining concentration of ClO⁻ in solution = 0.015 - 0.000185

                                                                                     = 0.0132            

moles of HClO = 0.0175 + 0.00185

                         = 0.0194

Using Henderson equation pH = Pka + log

                                                    = 7.52 + log

                                                     = 7.35

Part C

         pH after addition of 85 mg of HBr

moles of NaOH    

So the remaining concentration of ClO⁻ in solution = 0.015 + 0.00213

                                                                                     = 0.0171            

Moles of concentration of ClO⁻ = 0.171(M)

moles of HClO = 0.0175 - 0.00213

                         = 0.0154

Moles in 100 ml Buffer = 0.154(M)

Using Henderson equation pH = Pka + log

                                                    = 7.52 + log

                                                     = 7.56

it should be p1 v1 = p2 v2

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation: