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ph of strong acid = 4
[H+] = 10-4 = molarity of acid
in 0.1L , moles of acid = 0.1* 10-4
ph of base =10
[H+]=10-10
[OH-]=10-14/10-10=10-4 = molarity of base
in 0.2L , moles of acid = 0.2*10-4
now nutralization will occur and acid is completely nutralised by base , at the end base remaining due to higher concentration....
now final molariry = remaining moles/total vol = (0.2*10-4-0.1*10-4 ) /0.3
concentration of base =10-4/3
[H+] = 3*10-14/10-4 =3*10-10
ph=-loh[H+] = 10-log3= 9.523
3.
NaOH reacts with HCl : NaOH + HCl → NaCl + H2O 1 mol NaOH reacts with 1 mol HCl Mol HCl in 9.9mL of 1.0M solution = 9.9/1000*1.0 = 0.0099 mol HCl Mol NaOH in 100mL of 0.1M solution = 100/1000*0.1 = 0.01 mol NaOH On mixing the 0.0099 mol HCl will be neutralised and 0.01 - 0.0099 = 1*10^-4 mol NaOH dissolved in 109.9mL solution Molarity of NaOH solution = (1*10^-4) / 0.1099 = 9.099*10^-4M pOH = -log ( 9.099*10^-4) pOH = 3.04 pH = 14.00 - pOH pH = 14.00 - 3.04 pH = 10.96
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