- log(1 M propanoic acid) = 0 pH ========( remember, acids may register below scale )
pH = 0 for hydrochloric acid (1 M HCl)
- log(0.12 M) = 0.92 pH ---------------
- log(0.900 M cyanic acid) = 0.05 pH =============very strong acid!
2.4
- log(1 M propanoic acid) = 0 pH ========( remember, acids may register below scale )
pH = 0 for hydrochloric acid (1 M HCl)
- log(0.12 M) = 0.92 pH ---------------
- log(0.900 M cyanic acid) = 0.05 pH =============very strong acid!
Propanoic acid CH3CH2COOH is a weak acid. pKa=4.89 => Ka=1.3*10-5Use the (almost correct approximation) formula:[H+] = SQRT [ Ka * ca] , if at least [Ka/ca]< 0.04 and pH< 6.8 (which is true in this case)So: pH = 0.5*pKa - 0.5*log(ca) = 0.5*4.89 - 0.5*log(0.265) =2.445-(-0.288) = 2.733 = 2.7
2.4
pH values depend on the concentration of acidsolutionfor eg.0.02 M HF has pH=2.47
It depends on just how dilute it is. If we're talking about a strong acid - in this case, acids like perchloric (HClO4), sulfuric (H2SO4), or nitric (HNO3) - and assume that the acid dissociates completely (an ideal situation; in sulfuric acid's case, only the first proton is assumed to dissociate), the pH of the acid is the negative logarithm of its molarity. For example, if you have 1 × 10-3 M (.001 mol · L-1) hydrochloric acid (HCl), its pH will be -log(1 × 10-3) = 3. (Since real life is a non-ideal situation it will actually be slightly higher, but we can disregard that.) That's the easy part and only applies to strong acids. For other (weak) acids of formula HA ⇌ H+ + A-, the pH is dependent upon the acid dissociation constant pKa, in which case pH = pKa + log([A-]/[HA]). Say you have a weak acid with a pKa of 2.0 and a molarity of .01 M. Since pKa = -log(Ka), that means that Ka = .01. The definition of Ka is [A-][H+]/[HA]. Let's call [A-] and [H+] x for this purpose; this makes [HA] = .01 - x; thus, .01 = x2/(.01 - x). Solving for x gives x2 + .01x - .0001 = 0; using the quadratic formula we get .00618 M. Now we may derive the pH. pH = pKa + log([A-]/[HA]) = 2 + log(.00618/.00382) = 2.21.
The following table is extracted from the Internet:* For pH=3: mix 982,3 mL 0,1 M acetic acid with 17,7 mL 0,1 M sodium acetate* For pH=4: mix 847,0 mL 0,1 M acetic acid with 153 mL 0,1 M sodium acetate* For pH=5: mix 357 mL 0,1 M acetic acid with 643 mL 0,1 M sodium acetate* For pH=3: mix 52,2 mL 0,1 M acetic acid with 947,8 mL 0,1 M sodium acetate
Yes, 0.1 M HCl has pH value of 1.0, 1.0 M HCl has pH value of 0.0, 10.0 M pH = -1.0 (even negative is possible, but this is about the limit)
D) 5
Since HCl is a strong acid it completely dissociates. Therefore [H+] = [HCl] and this case = 0.25 M. pH = -log [H+] = 0.602