The dissociation constant, Ka, for HCN(aq) is 6.2 Times 10^-10. What is the pH o

Question

The dissociation constant, Ka, for HCN(aq) is 6.2 Times 10^-10. What is the pH of a 0.10 molar solution of sodium cyanide? 5.10 8.90 9.21 11.10 11.30 A mixture contains 0.0600 moles of NaH_2P0_4 and 0.0800 moles of K_2HP0_4. It is titrated with 0.500 molar NaOH(aq) to neutralize it completely. How many mL of the NaOH solution are required? 280. mL 313.3 mL 372 mL 400. mL 440. mL A solution is made by taking 100.0 mL of 2.0 Times 10^-2 molar NaOH(aq) and mixing in 50.0 mL of 3.0 Times 10^-2 molar HC1(aq). The acidity of a small portion was tested using tyrolian blue indicator, for which the pK_ind = 12.5, and the colors are blue for Hind, and yellow for lnd^-. what color would be observed as a result of the testing? blue green yellow orange red For H_3PO_3, which is actually a diprotic acid, Ka_1 = 1.0 Times 10^-2 and Ka_2 = 2.6 Times 10^-7. Calculate a value for the [H+] in a 0.500 molar solution of H_3P0_3. 1.0 Times 10^-2 mol L^-1 6.6 Times 10^-2 mol L^-1 7.1 Times 10^-2 mol L^-1 8.5 Times 10^-3 mol L^-1 3.3 Times 10^-3 mol L^-1 A solution is made by taking 2.42 grams of NaCl, 6.44 grams of NaCl0_4, 3.81 grams of KN0_3, and 4.64 grams of KBr in enough water to make 500 mL of solution. A portion was treated with thymol blue indicator, for which the pK_ind =8.8, and the colors are yellow for Hind, and blue for Ind". What color should be obtained with the test portion? blue green yellow orange red

Explanation / Answer

HCN = H+ and CN-

Ka = [H+][CN-]/[HCN]

Ka = 6.2*10^-10

then

[H+]= [CN-] = x

[HCN] = 0.1-x

Ka = [H+][CN-]/[HCN]

6.2*10^-10 = (x*x)/(0.1-x)

x = H+ = 7.87*10^-6

pH = -log(7.87*10^-6) = 5.104

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