Using the quadratic equation to calculate [H_3O+] in 0.00250 M HNO_2, what are t
ID: 1019323 • Letter: U
Question
Using the quadratic equation to calculate [H_3O+] in 0.00250 M HNO_2, what are the values of a, b, c and x, where a, b, and c are the coefficients in the quadratic equation ax^2 + bx + c = 0, and x is [H_3O+]? Recall that K_a = 4.5 times 10^-4. Express a, b, c, and x numerically separated by commas. Using the method of successive approximations, what is [H_3O+] in 0.500 M HC1O_2 (K_a = 1.1 times 10^2) and how many iterations (n) are needed to validate that a constant value had been obtained to two significant figures? Express [H_3 O+] using two significant figures, followed by a comma and the number of iterations as an integer. Using the method of successive approximations, what is [H_3O+] in 0.00500 M CH_2ClCOOH(aq) (K_a = 1.4 times 10^3) and how many iterations (n) are needed to validate that a constant value had been obtained to two significant figures? Express [H_3O+] using two significant figures, followed by a comma and the number of iterations as an integer.Explanation / Answer
HNO2 +H2O-------> H3O+ + No2-
I 0.0025 0 0
C -x +x +x
E 0.0025-x +x +x
Ka = [H3O+][No2-]/[HNO2]
4.5*10-4 = x*x/0.0025-x
4.5*10-4 (0.0025-x) = x2
x = 00086
[H3O+] = x =0.00086M
HClO2 + H2O --------> H3O+ + ClO2-
I 0.5 0 0
C -x +x +x
E 0.5-x +x +x
Ka = [H3O+][ClO2-]/[HClO2]
1.1*10-2 = x*x/0.5-x
1.1*10-2 *(0.5-x) = x2
x = 0.068
[H3O+] = x = 0.068M
c.
CH2ClCOOH + H2O ---------> H3O+ + CH2ClCOO-
I 0.005 0 0
C -x +x +x
E 0.005-x +x +x
Ka = [H3O+][CH2ClCOO-]/[CH2ClCOOH]
1.4*10-3 = x*x/0.005-x
1.4*10-3*(0.005-x) = x2
x = 0.002
[H3O+] = x = 0.002M
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