- log(0.001 M NH4Cl)
= 3 pH
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3. since the [H+]=0.001 M then pH= -log[H+] -log(0.001)=3 pH=3.
This represent a buffer of a weak base (NH3) and the conjugate acid (NH4+), so one can use a form of the Henderson Hasselbalch equation like pOH = pKb + log [conj.acid][base]. The pKb for NH3 is 9.25, so pH = 9.25 + log [0.17]/[0.13] = 9.25 + 0.12 = 9.37 = pH
NH4Cl-------------ammonium chloride
-log(0.1 M) = 1 pH
pH=1
3. since the [H+]=0.001 M then pH= -log[H+] -log(0.001)=3 pH=3.
5
This represent a buffer of a weak base (NH3) and the conjugate acid (NH4+), so one can use a form of the Henderson Hasselbalch equation like pOH = pKb + log [conj.acid][base]. The pKb for NH3 is 9.25, so pH = 9.25 + log [0.17]/[0.13] = 9.25 + 0.12 = 9.37 = pH
NH4Cl-------------ammonium chloride
-log(0.1 M) = 1 pH
pH=1
- log(0.12 M) = 0.92 pH ---------------
-log(0.5 M HF) = 0.3 pH
- log(0.000626 M H2SO4) = 3.2 pH -----------
there are 1 cm^2 = .0001 M^2
If you mean 0.001, the prefix "milli" (abbreviated "m") is used for that.
- log(0.00450 M HCl)= 2.3 pH=======