pH=−log10[H+] [H+]=10−pH [H+] of solution 1=10−3 [H+] of solution 2=10−4 [H+] of solution 3=10−5 The volume taken in each case is 1L =10−3(1+1×10−1+1×10−2. =10−3(‌
1
1
+‌
1
10
+‌
1
100
) =10−3(‌
111
100
)=1.11×10−3 Therefore, H+ion concentration in mixture of equal volume of these acid solutions =‌