What Is the pH of a 0.1300 M HCN Solution?

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The discussion focuses on calculating the pH of a 0.1300 M HCN solution, a weak acid with a dissociation constant (Ka) of 4.9 x 10^-10. The equilibrium expression for the dissociation of HCN is established, leading to the calculation of hydronium ion concentration ([H3O+]) using the formula derived from Ka. The calculated [H3O+] is approximately 8.54 x 10^-6, resulting in a pH of about 5.07. Participants confirm the setup of the calculations, noting that the volume change from adding HCN is negligible. Overall, the pH calculation appears correct, with emphasis on significant figures and assumptions made during the process.
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Hydrocyanic acid is a weak acid (Ka = 4.9 x 10-10). If 0.1300 moles of gaseous HCN are dissolved in 0.8700 liters of water. Determine the pH of the HCN solution formed.

HCN + H2O <-> H3O+ + CN-

K_a = [CN-][H3O+]/[HCN]

M HCN = 0.1300 mol HCN/0.87351338 L = 0.14882428 M HCN ??

For L solution: 0.1300 mol HCN*(27.026 g/1 mol HCN) = 3.51338 g HCN

870 g H2O + 3.51338 g HCN = 873.51338 = 873.51338 mL = 0.87351338 L??

K_a = [x][x]/[0.14882428 - x]

Assuming 0.14882428 - x = 0.14882428,
x = sqrt(4.9 x 10-10*0.14882428) = 8.539549E-6 = [H3O+]

pH = -log(8.539549E-6) = 5.068565065 = 5.07 ?

Thanks.
 
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set up seems fine to me
 
Soaring Crane said:
870 g H2O + 3.51338 g HCN = 873.51338 = 873.51338 mL = 0.87351338 L??

You have only two significant digits in the Ka, thus you may safely assume concentration is 0.13/0.87M - and neglect volume change.

Rest is OK.
 
870 g H2O + 3.51338 g HCN = 873.51338 = 873.51338 mL = 0.87351338 L


That doesn't make a big difference, but it is not right. Just assume total V is .870 L .
 
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