what is the pH of 0.00001m solution?

To calculate the pH of a 0.00001 M solution, you need to know the dissociation constant (Ka) of the acid in question, as well as the chemical equation for its dissociation. If you don't have this information, you cannot accurately calculate the pH of the solution. Assuming you are asking about a weak acid, you can use the following equation to calculate the pH: pH = pKa + log([A-]/[HA]) Where pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid. For example, let's say the weak acid in question is acetic acid (CH3COOH), which has a pKa of 4.76. The dissociation of acetic acid can be represented by the following equation: CH3COOH + H2O ⇌ CH3COO- + H3O+ The equilibrium expression for this reaction is: Ka = [CH3COO-][H3O+]/[CH3COOH] If the concentration of acetic acid is 0.00001 M, then the concentrations of CH3COO- and H3O+ at equilibrium can be assumed to be very small compared to [CH3COOH], so we can make the approximation that [CH3COO-][H3O+] ≈ 0. Therefore, we can simplify the equilibrium expression to: Ka = [CH3COO-][H3O+]/[CH3COOH] ≈ [CH3COO-][H3O+]/0.00001 Assuming that [CH3COO-] ≈ [H3O+], we can write: Ka = [H3O+]^2/0.00001 [H3O+]^2 = Ka x 0.00001 [H3O+] = sqrt(Ka x 0.00001) [H3O+] = sqrt(1.8 x 10^-5 x 0.00001) [H3O+] = 0.00134 M Now we can use the pH equation: pH = pKa + log([A-]/[HA]) pH = 4.76 + log(0.00134/0.00001) pH = 4.76 + 2.13 pH = 6.89 Therefore, the pH of a 0.00001 M solution of acetic acid is approximately 6.89.