A 50.0 ml sample of 0.155 M HNO_2(aq) is titrated with 0.100 M NaOH(aq). What is

Question

A 50.0 ml sample of 0.155 M HNO_2(aq) is titrated with 0.100 M NaOH(aq). What is the addition of 25.0 mL of NaOH? (Ka of HNO_2 = 4.5 Times 10^-4) 3.02 3.22 3.67 3.86 4.05 What is the maximum hydroxide-ion concentration that a 0.019 M MgCl_2 solution could have without causing the precipitation of Mg(OH)^2? For Mg(OH)_2, K_sp = 1.8 Times 10^-11. 4.2 Times 10^-6 1.7 Times 10^-4 1.2 Times 10^-8 9.5 Times 10^-9 3.1 Times 10^-5 If 25 mL of 0.750 M HCL are added to 100, mL of 0.283 M NaOH, what is the final pH? 12.88 1.12 13.35 0.65 What is the pH a buffer composed of 0.35 M H_2PO^-_4(aq) and 0,65 M HPO^2-_4(aq)?(K_a of H_2PO^-_4 is 6.2 Times 10^-8) 6.94 7.21 7.48 7.73 9.06

Explanation / Answer

53. [HNO2] remain = (0.155 M x 50 ml - 0.1 M x 25 ml)/75 ml = 0.07 M

[NO2-] formed = 0.1 M x 25 ml/75 ml = 0.033 M

pKa = 3.34

pH = pKa + log(A-/HA)

     = 3.34 + log(0.033/0.07)

     = 3.02

Answer : a. 3.02

55. [OH-] = sq.rt.(Ksp/[Mg2+]) = sq.rt.(1.8 x 10^-11/0.019) = 3.1 x 10^-5 M

Answer : e. 3.1 x 10^-5

57. [H+] = 0.75 Mx 25 ml = 18.75 mmol

[OH-] = 0.28 m x 100 ml = 28 mmol

excess [OH-] = 0.074 M

pOH = -log[OH-] = 1.13

pH = 14 - pOH = 12.88

Answer : a. 12.88

59. pH = 7.2 + log(0.65/0.35) = 7.48

Answer : c. 7.48

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