HCl(aq) + KOH(aq) ----------------> KCl(aq) + H2O(l)
1 mole 1 mole
HCl KOH
M1 = M2 = 4.96*10^-2M
V1 = 11.9ml V2 = 35.3ml
n1 = 1 n2 = 1
M1V1/n1 = M2V2/n2
M1 = M2V2n1/V1n2
= 4.96*10^-2 *35.3*1/11.9*1 = 0.147M
The molarity of HCl = 0.147M
part-B
2HClO4(aq) + Ba(OH)2(aq) -----------------> Ba(ClO4)2(aq) + 2H2O(l)
2 mole 1mole
HClO4 Ba(OH)2
M1= 0.326M M2 =
V1 = 15.7ml V2 = 30.1ml
n1 = 2 n2 = 1
M1V1/n1 = M2V2/n2
M2 = M1V1n2/V2n1
= 0.326*15.7*1/(30.1*2)
= 0.085 M
The molarity of Ba(OH)2 = 0.085M
part-C
PH = PKa + log[CN^-]/[HCN]
9.12 = 9.4 + log[CN^-]/[HCN]
log[CN^-]/[HCN] = 9.12-9.4
log[CN^-]/[HCN] = -0.28
[CN^-]/[HCN] = 0.525/1
part-D
PH = PKa + log[HCO3^-]/[H2CO3]
6.18 = 6.38 + log[HCO3^-]/[H2CO3]
log[HCO3^-]/[H2CO3] = 6.18-6.38
[HCO3^-]/[H2CO3] = 0.63/1
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