Find the pH and percent ionization of each HF solu...

Question

Find the pH and percent ionization of each HF solution. (Ka for HF is 6.8 x 10-4.)Find the percent dissociation of a 0.270 M HF solution.

Jules Bruno · Accepted Answer

Recall that the percent ionization is given by:% ionization=[H+][HA]initial×100The dissociation of HA is as follows:HF(aq) + H2O(l) ⇌ H3O+(aq) + F–(aq); Ka = 6.8 × 10–4[readmore]The ICE table and Ka expression.The Ka expression for HF is:Ka=productsreactants=[H3O+][F-][HF]6.8×10-4=[H3O+][F-][HF]6.8×10-4=(x)(x)0.270-x[HF]Ka=0.2706.8×10-4=397.1 < 500x20.270-x=6.8×10-4x2=(6.8×10-4)(0.270-x)x2=1.836×10-4-6.8×10-4xx2+6.8×10-4x-1.836×10-4=0  ax2 +           bx                  +                  c                  = 0 The quadratic formula is:x=-b±b2-4ac2aFrom the equation, we have a = 1, b = 6.8 × 10–4, and c = – 1.836 × 10–4. x=-6.8×10-4±(6.8×10-4)2-4(1)(-1.836×10-4)2(1)x=-6.8×10-4±4.624×10-7-(-7.344×10-4)2x=-6.8×10-4±4.624×10-7+7.344×10-42x=-6.8×10-4±7.348624×10-42x=-6.8×10-4±0.02712The calculation then splits into two ways:x=-6.8×10-4+0.02712x=0.026422                       x= 0.01321 M  x=-6.8×10-4-0.02712x=-0.027782   x=– 0.01389 MThis means [H3O+] = x = 0.01321 M. We can now calculate for percent dissociation:% ionization=0.01321 M0.270 M×100% % ionization = 4.89 %The percent dissociation of 0.270 M HF solution is 4.89 %.

Problem: Find the pH and percent ionization of each HF solution. (Ka for HF is 6.8 x 10-4.)Find the percent dissociation of a 0.270 M HF solution.

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Problem: Find the pH and percent ionization of each HF solution. (Ka for HF is 6.8 x 10-4.)Find the percent dissociation of a 0.270 M HF solution.

FREE Expert Solution

Problem Details

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