The pH is defined as the minus logarithm of concentration of H+ ions.
Since, HCl dissociates completely in water, the concentration of H+ ions will be 10M.
Therefore, pH = -log [H+] = -log 10 = -1.
when temperature is increased of a reaction by 10c ,then rate of reaction increases due to ?
a) net increase in Ea
b)increases in number of collisions
c) increases in fruitful collisions of reactants
d) Decreases in Ea
Q2;
AT which point heat absored without change in phaes ?
a) Melting point
b) boiling point
c) both
d) none
Q3;
London dispersion foreces exist in ;
a) in bromine water
b) in hcl
c) in molecules of Ne gas
d) in molecules of CH3CL
Because HCl is a strong acid it will dissociate completely in water to form hydronium ions. The pH can therefore be calculated by taking the negative log of the concentration.
pH=-log10
=-1
Therefore the pH is negative one. This indicates that it is a very strong acid. This value is also applicable even though it is outside of the standard scale.
4nM HCl = 0.004 M
-log(0.004 M)
= 2.4 pH
2
3
Making 25 mm Borate HCL buffer with a pH of 8.8 will require careful calculation and measurement. The pH can be easily adjusted after formulation. Temperature can impact the pH.
- log(0.00450 M HCl)= 2.3 pH=======
Remember pH = -(log(10)[H^+] Substituting pH = -(log(10)[5 x 10^(-4)] On the calculator pH - -(-3.30102996) pH = 3.30 ( to 2 d.p.).
- log(0.25 M HCl) = 0.6 pH ------------
Most probably PH 12, that is if you add 8+4
Making 25 mm Borate HCL buffer with a pH of 8.8 will require careful calculation and measurement. The pH can be easily adjusted after formulation. Temperature can impact the pH.
0.1 M NaCl10 mM Tris-HCl (pH 8.0)1 mM EDTA (pH 8.0)
- log(0.00450 M HCl)= 2.3 pH=======
137 mM NaCl, 25 mM Tris-HCl [pH 7.4], 0.7 mM Na2HPO4, 5 mM KCl
Remember pH = -(log(10)[H^+] Substituting pH = -(log(10)[5 x 10^(-4)] On the calculator pH - -(-3.30102996) pH = 3.30 ( to 2 d.p.).
Tartaric acid - 18.75 mM Sodium Tartrate Dihydrate - 6.25 mM pH of this mixture should be near 2.9, adjust pH down to 2.5 using HCl, and QS to 1 L by water.
- log(0.25 M HCl) = 0.6 pH ------------
Most probably PH 12, that is if you add 8+4
.260 M of HCL, not 260 More than likely correct, but, - log(0.260 M HCl) = 0.6 pH ----------- ( pH can be below 1 )
if its complete dissociation, then the products would be a salt and water, which means the pH is 7 or neutral. OMG, if the pH is currently 4 then [H+] = 1.0 e-4 M (pH = -log[H+]) if you add 0.003 moles then 1.0e-4 M +.003 M = .0031 M (Since the strong acid HCL completely dissociates in aq solution) pH = -log [.0031M] = 2.51
pH=-lg[H+][H+]=10-pHWith pH=2.0:Corrected:[H+]= 10-pH = 10-2.0 = 0.010 M HCL
Since HCl is a strong acid it completely dissociates. Therefore [H+] = [HCl] and this case = 0.25 M. pH = -log [H+] = 0.602