Chemistry Help Please

[Belgarion]

I cant get these questions, any answers would be appreciated:

1. HCOOH and CH3COOH have Ka vaulues iof 0.00018 M and 0.000017 M respectively.

calculate the pH of 0.01 M of HCOOH

calculate the concentration of CH3COOH need to produce same pH as first part


2. At 5 degrees, the concentration of OH ions in a pure water sample is 0.00000004 M

calculate concentration of hydronium ions

calculate value of Kw

Calculate pH of pure water


3. CO2(g) + 2H20(l) double arrows H30+ + HCO3-

Concentration of C02 is 0.000013 M and pH is 7.4

calculate concentraion of H30+ ions

Calculate Ka assuming [H30+]=[HCO3-]



4. A certain weak acid has a Ka of 0.00002 M.     HA + H20 double arrows H30+ A-

Calculate pH of 1M solution of acid

Small amount of solid salt with formula NaA is added until it dissolves. Concentration of A- ions is 0.0021M. Calculate new pH

[Jenny_2108]

1. Ka(HCOOH)=0.00018
    Ka=[H+]^2/[HCOOH] => [H+]=0.001341
    pH=-log(0.001341]=2.87

    [H+]^2/[CH3COOH]=0.000017 => [CH3COOH]=1.8*(10^-6)/0.000017=0.106M

3. pH=7.4 => [H+]=10^-7.4
    Ka=[H+]^2/[CO2]=0.003

4. Ka=[H+]^2/[HA]=0.00002M => [H+]=0.00447
    pH=-log(0.00470=2.35

    [A-]=0.0021M
    Ka=[H+][A-]/[HA]=0.00002M
    [H+]= 0.00002*1/0.0021=0.0095M
    New pH=-log(0.0095)=2.02

2. At 5 degrees, the concentration of OH ions in a pure water sample is 0.00000004 M

calculate concentration of hydronium ions

calculate value of Kw

Calculate pH of pure water

Is it at 5 or 25 degrees? Because at 5 degrees, we cant use the formula [H+][OH-]=10^-14 

   

[Belgarion]

its at 5 degrees

[Jenny_2108]

its at 5 degrees

In the textbook, its said at 5 degrees, pH=7.37 and Kw=1.85*10^-15

[golden]

its at 5 degrees

In the textbook, its said at 5 degrees, pH=7.37 and Kw=1.85*10^-15

Because this is a neutral solution, you can find [H+], because it is the same as [OH-]. Kw is just a multiplication away!