Use the Goal Seek spreadsheet at the end of the chapter to find the pH of 1.00 L of solution containing 0.030 mol HA ([latex]pK_{a}[/latex] = 2.50) and 0.015 mol NaA. What would the pH be with the approximations [HA] = 0.030 and [latex][A^{−}][/latex] = 0.015?

We use Goal Seek to vary cell B5 until cell D4 is equal to [latex]K_{a}[/latex]. The spreadsheet shows that [latex][H^{+}] = 4.236 × 10^{−3}[/latex] (cell B5) and pH = 2.37 in cell B7. A B C D 1 Ka = 10^-pKa = 0.00316228 Reaction quotient 2 Kw = 1.00E-14 for Ka = 3 FHA = 0.03 [H+][A-]/[HA] = 4 FA = 0.015 0.0031623 5 H = 4.236E-03 <-Goal Seek solution 6 OH = Kw/H = 2.3609E-12 D4 = H*(FA+H-OH)/ 7 pH = -logH = 2.37 (FHA-H+OH) If we were doing this problem by hand with the approximation that what we mix is what we get, [latex][H^{+}] = K_{a}[HA]/[A^{−}] = 10^{−2.50}[0.030]/[0.015][/latex] = 0.006 32 M ⇒ pH = 2.20.

Question 8.E.L: Use the Goal Seek spreadsheet at the end of the chapter to f...

Quantitative Chemical Analysis

Use the Goal Seek spreadsheet at the end of the chapter to find the pH of 1.00 L of solution containing 0.030 mol HA ( $pK_{a}$ = 2.50) and 0.015 mol NaA. What would the pH be with the approximations [HA] = 0.030 and $[A^{−}]$ = 0.015?

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We use Goal Seek to vary cell B5 until cell D4 is equal to $K_{a}$ .
The spreadsheet shows that $[H^{+}] = 4.236 × 10^{−3}$ (cell B5) and pH = 2.37 in cell B7.

	A	B	C	D
1	Ka = 10^-pKa =	0.00316228		Reaction quotient
2	Kw =	1.00E-14		for Ka =
3	FHA =	0.03		[H+][A-]/[HA] =
4	FA =	0.015		0.0031623
5	H =	4.236E-03	<-Goal Seek solution
6	OH = Kw/H =	2.3609E-12		D4 = H*(FA+H-OH)/
7	pH = -logH =	2.37		(FHA-H+OH)

If we were doing this problem by hand with the approximation that what we mix is what we get, $[H^{+}] = K_{a}[HA]/[A^{−}] = 10^{−2.50}[0.030]/[0.015]$ = 0.006 32 M ⇒ pH = 2.20.