the plH of the solution if 200. mL of 0.70 M HCN is titrated with 0.10 mol KOH.
ID: 304394 • Letter: T
Question
the plH of the solution if 200. mL of 0.70 M HCN is titrated with 0.10 mol KOH. 3. Find the pH of a solution in which 500. mL of 0.20 M NH.Cl is titrated with 0.10 mol HNO.. pK. (NH)-9.25 pK, (HNO,) --1.4 4. Find the phi of a solution when 120. g of NaOH is added to 3.0 L of 0.75 M HF. Assume no change in volume. pK (HF) 3.17 pH = 5. Find the pH of a solution when 100. mL of 0.60 M HC H,0, is combined with 100.ml of 0.07 M NaCaHjO pK, (HC2H,0:) 4.76 pH = 6. Find the pH of the solution if 15 g of NaNO, is added to 300.ml of 0.10 M HNO, Assume no change in volume pk. (HNO) = 3.40 7. Find the pH of a solution in which 1.25 L of 0.50 M NaCIO2 and 0.75 L of 0.80 M are combined, followed by 0.30 mol of HCL pK, (HCO) # 1.96 pK, (HCI)-7 pH-Explanation / Answer
2) no of mol of HCN = 0.2*0.7 = 1.4 mmol
no of mol of KOH added = 0.1 mol
pH = pka + log(KCN/HCN)
= 9.21+log(0.1/(1.4-0.1))
= 8.1
3) HNO3 is strong acid,NH4Cl weakly acidic.so that, acidity depends majorly on HNO3
concentration of HNO3 = n/v = 0.1/0.5 = 0.2 M
pH = -log(H3O+)
= -log(0.2)
= 0.7
4) No of mol of HF taken = M*V
= 0.75*3
= 2.25 mol
No of mol of NaOH = 120/40 = 3 mol
so that, total HF reactes with NaOH.and some extra NaOH is left in the mixture.
pH depends on extra NaOH
concentration of extra NaOH = n/v = (3-2.25)/3 = 0.25 M
pOH = - log(OH-)
= -log0.25
= 0.6
pH = 14-0.6 = 13.4
5) No of mol of CH3COOH = 0.1*0.6 = 0.06 mol
No of mol of CH3COONa = 0.1*0.07 = 0.007 mol
pH = pka + log(CH3COONa/CH3COOH)
pka of ch3cooh = 4.76
pH = 4.76+log(0.007/0.06) = 3.83
6) No of mol of NaNO2 = 15/69 = 0.217 mol
No of mol of HNO2 = 0.3*0.1 = 0.03 mol
pka of HNO2 = 3.4
pH = 3.4+log(0.217/0.03) = 4.26
7) No of mol of HClO2 = 0.75*0.8 = 0.6 mol
No of mol of nAcLo2 = 1.25*0.5 = 0.625 mol
PKA OF hcLo2 = 1.96
pH = pka + log(NaClO2-HCl)/(HClO2+HCl)
= 1.96+log((0.625-0.3)/(0.6+0.3))
= 1.52
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