ph question ?

A.
pH= log(1/[H+])
pH = -log[H⁺]
log[H⁺] = -pH
Hence, [H⁺] = 10^(-pH) M

[H⁺] of the coffee = 10⁻⁵ M = 1.00 × 10⁻⁵ M
[H⁺] of the milk = 10⁻⁶˙⁶ M = 2.51 × 10⁻⁷ M

For the solution containing half coffee and half milk:
[H⁺] due to the coffee = (1.00 × 10⁻⁵ M) × (1/2) = 5.00 × 10⁻⁶ M
[H⁺] due to the milk = (2.51 × 10⁻⁷ M) × (1/2) = 1.26 × 10⁻⁷ M
Total [H⁺] in the mixture = (5.00 × 10⁻⁶ + 1.26 × 10⁻⁷) M = 5.13 × 10⁻⁶ M
pH of the mixture = log{1/(5.13 × 10⁻⁶)} = 5.29


B.
Assume that a mixture of y% of coffee and (100 - y)% of milk has a pH of 5.51.

In the mixture:
[H⁺] due to the coffee = (1.00 × 10⁻⁵ M) × y%
[H⁺] due to the milk = (2.51 × 10⁻⁷ M) × (100 - y)%

Total [H⁺] (in M) in the mixture:
(1.00 × 10⁻⁵) × y% + (2.51 × 10⁻⁷) × (100 - y)% = 10⁻⁵˙⁵¹
(1.00 × 10⁻⁵) × y% + (2.51 × 10⁻⁷) × (100 - y)% = 3.09 × 10⁻⁶
(1.00 × 10⁻⁷)y + (2.51 × 10⁻⁷) - (2.51 × 10⁻⁹)y = 3.09 × 10⁻⁶
(9.746 × 10⁻⁸)y = 2.839 × 10⁻⁶
y = 29.1
100 - y = 70.9

The mixture contains 29.1% of coffee and 70.9 % of milk.

Concentration of H+ in the coffee is 10^(-5); in the milk it's 2.512 x 10^(-7). In the mixture, it will be
(1/2)[10^(-5) + 2.512*10^(-7)], or
5.13 x 10^(-6). The new pH is
-[log(5.13) - 6]
= 5.29.

Part (b).
Concentration in the pH 5.51 mixture would be 10^(-5.51) = 3.09 x 10^(-6). So you want
x*10^(-5) + (1-x)*2.512*10^(-7) = 5.51,
where "x" is the coffee fraction in the new mixture. I leave it to you to finish the problem. The answer should be less than half coffee...