The hydronium concentration ([H_2O^+]) in a sample of wine is 3.2 times 10^-4 M.
ID: 527134 • Letter: T
Question
The hydronium concentration ([H_2O^+]) in a sample of wine is 3.2 times 10^-4 M. What is the pH of the wine sample? 3.2 3.5 4.0 7.0 A 0.10 M solution of the deadly poison hydrogen cyanide, HCN, has a pH of 5.2. Calculate the [H_3O^+] of the solution 0.794 M H_3O^+ 5.2 M H_3O^+ 6.31 times 10^-4 M H_2O^- 1.0 times 10^-13 M H_3O^+ A 0.10 M solution of ammonium ion, NH_4^+, has a pH of 5.6. Ammonium ion is a strong acid. Weak acid. Strong base. Weak base. The OH^- concentration in a milk sample 6.2 times 10^-8 M. The value of [H_3 O^+] in the milk sample is 6.2 times 10-8 M and the milk is basic. 1.0 times 10-7 M and the milk is neutral. 6.2 times 10-7 M M and the milk is acidic 1.6 times 10-7 M M and the milk is acidic A cleaning solution is found to have [OH^-] = 5.0 times 10^-3 M. What is the pH? 2.30 5.00 11.70 12.00 When a 10.00 mL sample of household vinegar (dilute aqueous acetic acid) is titrated, 25.0 ml. of 0.200 M NaOH solution is required to reach the end point. What is the acid concentration of the vinegar in moles per liter. The neutralization reaction is CH_3CO_2H(aq) + NaOH(aq) rightarrow CH_3CO_2^-Na^+(aq) + H_2O(l) 0.890 M 0.500 M 0.200 M 2.00 MExplanation / Answer
1. pH =-log [H3O+]= -log (3.2*10-4) =3.5 ( b is correct)
2. pH= 5.2, [H3O+] =10(-5.2)= 6.31*10-6 ( c is correct)
3. NH4+ ion is conjugate acid of NH3. NH3+ H2O ------>NH4+ + OH-
its pH= 5.6, lower pH values less than 7 suggests strong acid. and higher values of pH >7 suggest basicitiy.
So NH4+ is weak acid.
4. [OH-]= 6.2*10-8, pOH= -log [OH-] = 7.21, pH= 14-7.21= 6.79, [H3O+]= 10(-6.79)= 1.62*10-7 ( d is correct, milk is acidic since pH <7)
5, [OH]= 5*10-3, [H+] [OH-]= 10-14, [H+]= 10-14/ (5*10-3)= 2*10-12 , pH= 11.7, ( C is correct)
6. As per the reaction CH3COOH+ NaOH ------->CH3COONa + H2O
1 mole of NaOH neutralizes 1 mole of acetic acid.
moles of NaOH in 25ml of 0.2M= molarity* Volume in Liters= 0.2*25/1000
moles of acetic acid as well= 0.2*25/1000
volume of acetic acid given = 10ml= 10/1000
molarity of Acetic acid = 0.2*25/1000/10/1000 = 0.5M ( b is correct)
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